The Mean Value Theorem
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- The Theorem and Worked Example 1
- Worked example 2
- worked example 3
- worked example 4
- Exercise 1
- Exercise 2
- Exercise 3
- Exercise 4
- Exercise 5
- Exercise 6
- Exercise 7
- Exercise 8
- Exercise 9
- The Mean Value Theorem - a Variation
- Exercise 10
- Exercise 11
- Exercise 12
- Exercise 13
- Exercise 14
- Exercise 15
- Exercise 16
- Exercise 17
- Exercise 18
- Exercise 19
- Exercise 20
- Exercise 21
- Exercise 22
- Exercise 23
- Exercise 25
- Exercise 24
- Exercise 26
- Exercise 27

Rolle`s Theorem
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Summarizing Questions
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Advanced Exercises
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Darboux's Mean Value Theorem
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[{"Name":"The Mean Value Theorem","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Theorem and Worked Example 1","Duration":"17m 38s","ChapterTopicVideoID":8442,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8442.jpeg","UploadDate":"2019-11-14T07:19:53.6830000","DurationForVideoObject":"PT17M38S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.425","Text":"In this clip, I\u0027ll be talking about the mean value theorem in calculus,"},{"Start":"00:05.425 ","End":"00:07.450","Text":"which is attributed to Lagrange."},{"Start":"00:07.450 ","End":"00:11.630","Text":"In fact, it\u0027s often called Lagrange\u0027s mean value theorem."},{"Start":"00:11.630 ","End":"00:14.970","Text":"I\u0027m going to first of all state it abstractly and then"},{"Start":"00:14.970 ","End":"00:18.520","Text":"we\u0027ll understand it more practically."},{"Start":"00:18.520 ","End":"00:21.880","Text":"I\u0027ll tell you now that its main use for us will"},{"Start":"00:21.880 ","End":"00:25.360","Text":"be improving various inequalities although in general,"},{"Start":"00:25.360 ","End":"00:28.360","Text":"it has a lot more uses."},{"Start":"00:28.360 ","End":"00:35.400","Text":"Let me start formally by saying what it is, it\u0027s a theorem."},{"Start":"00:35.400 ","End":"00:37.500","Text":"Usually in a theorem is an if part and"},{"Start":"00:37.500 ","End":"00:40.590","Text":"a then part so that\u0027s what\u0027s going to be here also,"},{"Start":"00:40.590 ","End":"00:46.395","Text":"I\u0027m going to start off by saying if something and there\u0027s going to be 2 parts,"},{"Start":"00:46.395 ","End":"00:51.040","Text":"if 1 is true and 2 is true,"},{"Start":"00:51.040 ","End":"00:55.480","Text":"then I\u0027m going to get some conclusion."},{"Start":"00:55.480 ","End":"00:58.045","Text":"That\u0027s the general outline."},{"Start":"00:58.045 ","End":"01:02.300","Text":"If let\u0027s say it\u0027s a function"},{"Start":"01:03.800 ","End":"01:13.670","Text":"f and it satisfies 2 conditions, 1 and 2."},{"Start":"01:13.670 ","End":"01:16.410","Text":"Before I forget, let me write the and here."},{"Start":"01:16.410 ","End":"01:18.180","Text":"When I say here 1 and 2,"},{"Start":"01:18.180 ","End":"01:21.240","Text":"I mean and, I mean both."},{"Start":"01:21.240 ","End":"01:28.700","Text":"That f is continuous on the closed interval a,"},{"Start":"01:28.700 ","End":"01:30.560","Text":"b, and in case you\u0027re not sure what that is,"},{"Start":"01:30.560 ","End":"01:32.840","Text":"I\u0027ll tell you in a moment,"},{"Start":"01:32.840 ","End":"01:40.250","Text":"and that f is differentiable on the open interval a,"},{"Start":"01:40.250 ","End":"01:42.620","Text":"b, which we write like this."},{"Start":"01:42.620 ","End":"01:47.045","Text":"Now in case you\u0027re not familiar with this notation for intervals,"},{"Start":"01:47.045 ","End":"01:53.090","Text":"then this just means all the xs or any other letter you care to choose,"},{"Start":"01:53.090 ","End":"01:56.495","Text":"which are between a and b inclusive,"},{"Start":"01:56.495 ","End":"01:59.960","Text":"it\u0027s like saying all the numbers x such"},{"Start":"01:59.960 ","End":"02:03.875","Text":"that a is less than or equal to x, less than or equal to b."},{"Start":"02:03.875 ","End":"02:06.050","Text":"When we write round brackets,"},{"Start":"02:06.050 ","End":"02:08.154","Text":"we call it an open interval,"},{"Start":"02:08.154 ","End":"02:12.245","Text":"it means all the xs that are between a and b,"},{"Start":"02:12.245 ","End":"02:15.410","Text":"but not including the end points."},{"Start":"02:15.410 ","End":"02:18.364","Text":"Of course, in order for this to make sense,"},{"Start":"02:18.364 ","End":"02:21.360","Text":"a is less than b and so this is what we assume."},{"Start":"02:21.360 ","End":"02:24.585","Text":"A function satisfies these 2 things,"},{"Start":"02:24.585 ","End":"02:26.720","Text":"and I hope you remember what it means to be"},{"Start":"02:26.720 ","End":"02:29.990","Text":"continuous and what it means to be differentiable,"},{"Start":"02:29.990 ","End":"02:36.380","Text":"then there exists some point c between a and b,"},{"Start":"02:36.380 ","End":"02:43.405","Text":"c such that a is less than c is less than b."},{"Start":"02:43.405 ","End":"02:52.520","Text":"Such that the derivative of f at that point c is equal to"},{"Start":"02:52.520 ","End":"02:57.320","Text":"f of b minus f of"},{"Start":"02:57.320 ","End":"03:03.625","Text":"a over b minus a."},{"Start":"03:03.625 ","End":"03:08.170","Text":"This is what Lagrange\u0027s theorem says."},{"Start":"03:08.170 ","End":"03:09.610","Text":"Let me go over it again."},{"Start":"03:09.610 ","End":"03:13.345","Text":"We have a function which satisfies 2 conditions,"},{"Start":"03:13.345 ","End":"03:16.090","Text":"that f is continuous on the interval a,"},{"Start":"03:16.090 ","End":"03:19.255","Text":"b, from a to b inclusive,"},{"Start":"03:19.255 ","End":"03:24.535","Text":"which means that I can draw it without taking my pen off the paper so to speak,"},{"Start":"03:24.535 ","End":"03:27.670","Text":"and it\u0027s differentiable from a to"},{"Start":"03:27.670 ","End":"03:31.970","Text":"b. I don\u0027t care if it\u0027s differentiable or not at the end point."},{"Start":"03:31.970 ","End":"03:35.210","Text":"Given a function that satisfies these 2 conditions,"},{"Start":"03:35.210 ","End":"03:41.070","Text":"we\u0027re guaranteed that there exists some point c,"},{"Start":"03:41.070 ","End":"03:42.360","Text":"but there could be more than 1,"},{"Start":"03:42.360 ","End":"03:46.500","Text":"but there exists at least 1 such that c is between a and"},{"Start":"03:46.500 ","End":"03:52.010","Text":"b and which satisfies this equation that the derivative of f,"},{"Start":"03:52.010 ","End":"03:54.200","Text":"I mean it has a derivative is differentiable,"},{"Start":"03:54.200 ","End":"04:00.750","Text":"the derivative of f at this point c is exactly equal to this expression."},{"Start":"04:01.400 ","End":"04:04.705","Text":"There I\u0027ve presented it."},{"Start":"04:04.705 ","End":"04:08.360","Text":"Just to let you know what\u0027s coming up next."},{"Start":"04:08.360 ","End":"04:15.300","Text":"First, I\u0027m going to show you why this strange-looking formula is actually intuitive."},{"Start":"04:15.300 ","End":"04:18.775","Text":"Then I\u0027m going to explain how it\u0027s useful,"},{"Start":"04:18.775 ","End":"04:21.280","Text":"especially for solving inequalities."},{"Start":"04:21.280 ","End":"04:27.770","Text":"Then we\u0027re going to do some actual examples to really show that it\u0027s useful."},{"Start":"04:27.780 ","End":"04:34.940","Text":"Let\u0027s begin with the intuitive explanation of this theorem."},{"Start":"04:37.220 ","End":"04:44.364","Text":"In geometrical terms, what Lagrange\u0027s theorem tells us is as follows."},{"Start":"04:44.364 ","End":"04:49.900","Text":"If we have a function and that\u0027s this f of x in red,"},{"Start":"04:49.900 ","End":"04:52.795","Text":"which is defined between a and b,"},{"Start":"04:52.795 ","End":"04:55.340","Text":"and it\u0027s continuous from a to b,"},{"Start":"04:55.340 ","End":"05:00.120","Text":"and it also is differentiable between a and b,"},{"Start":"05:00.120 ","End":"05:02.710","Text":"possibly not at the end points."},{"Start":"05:02.710 ","End":"05:06.230","Text":"Then if I joined the secant,"},{"Start":"05:06.230 ","End":"05:07.400","Text":"don\u0027t worry about the word Secant,"},{"Start":"05:07.400 ","End":"05:11.420","Text":"if I join these 2 end points with a line segment,"},{"Start":"05:11.420 ","End":"05:16.730","Text":"then the slope of this is going to be the same as the slope of"},{"Start":"05:16.730 ","End":"05:22.325","Text":"the tangent line to some point c between a and b."},{"Start":"05:22.325 ","End":"05:30.470","Text":"In other words that\u0027s the point of contact where x equals c. The 2 blue slopes are equal."},{"Start":"05:30.470 ","End":"05:32.720","Text":"Let\u0027s see how that ties in with this."},{"Start":"05:32.720 ","End":"05:36.020","Text":"Let\u0027s take the 2 slopes and call them,"},{"Start":"05:36.020 ","End":"05:42.795","Text":"there\u0027s this 1 I\u0027ll call it the slope m from A to B,"},{"Start":"05:42.795 ","End":"05:47.175","Text":"that\u0027s the slope of the secant line as it\u0027s known."},{"Start":"05:47.175 ","End":"05:53.105","Text":"What that is equal to is the difference of the ys minus the difference of the xs."},{"Start":"05:53.105 ","End":"06:00.470","Text":"Now the y here is f of b because if the x coordinate here is b,"},{"Start":"06:00.470 ","End":"06:03.065","Text":"the y coordinate is f of b."},{"Start":"06:03.065 ","End":"06:06.700","Text":"Here this point would be a,"},{"Start":"06:06.700 ","End":"06:11.290","Text":"f of a. I could draw some dotted line."},{"Start":"06:13.760 ","End":"06:17.670","Text":"This is f of b and this is f of a."},{"Start":"06:17.670 ","End":"06:23.945","Text":"The difference in the ys is f of b minus f of a."},{"Start":"06:23.945 ","End":"06:28.450","Text":"Then the difference in the xs is b minus a."},{"Start":"06:28.450 ","End":"06:31.845","Text":"That gives us the slope of the secant."},{"Start":"06:31.845 ","End":"06:39.230","Text":"Now what about the slope of the tangent lines or call it m,"},{"Start":"06:39.230 ","End":"06:42.740","Text":"I don\u0027t know, of T for tangent."},{"Start":"06:42.740 ","End":"06:49.280","Text":"The slope of the tangent is the derivative of the function and we know it has"},{"Start":"06:49.280 ","End":"06:58.240","Text":"a derivative at the point c. Equating these 2,"},{"Start":"06:58.240 ","End":"06:59.630","Text":"which is what we\u0027ve done here,"},{"Start":"06:59.630 ","End":"07:04.840","Text":"just says that the slope of this tangent is the same as the slope of the secant."},{"Start":"07:04.840 ","End":"07:06.590","Text":"As we go along here,"},{"Start":"07:06.590 ","End":"07:14.290","Text":"drawing tangents at 1 point at least the tangent is going to be parallel to the secant."},{"Start":"07:14.290 ","End":"07:16.890","Text":"It actually could occur in more than 1 place,"},{"Start":"07:16.890 ","End":"07:18.620","Text":"if it\u0027s wavy curve,"},{"Start":"07:18.620 ","End":"07:19.760","Text":"it could be several."},{"Start":"07:19.760 ","End":"07:22.505","Text":"I could draw examples, but I won\u0027t."},{"Start":"07:22.505 ","End":"07:26.795","Text":"This is a geometrical interpretation of Lagrange\u0027s theorem."},{"Start":"07:26.795 ","End":"07:29.990","Text":"Now I\u0027d like to show you why this is useful,"},{"Start":"07:29.990 ","End":"07:32.555","Text":"at least why it\u0027s useful for us."},{"Start":"07:32.555 ","End":"07:35.959","Text":"I don\u0027t need this picture anymore."},{"Start":"07:35.959 ","End":"07:39.890","Text":"You remember I said something about inequalities,"},{"Start":"07:39.890 ","End":"07:41.735","Text":"all I see here is equality."},{"Start":"07:41.735 ","End":"07:47.240","Text":"But suppose that we know something about this f prime of"},{"Start":"07:47.240 ","End":"07:54.005","Text":"c. Suppose that I can deduce somehow an inequality about it."},{"Start":"07:54.005 ","End":"07:58.490","Text":"Say it\u0027s less than 1 thing and bigger than another thing,"},{"Start":"07:58.490 ","End":"08:01.524","Text":"doesn\u0027t matter what I\u0027ll symbolically say."},{"Start":"08:01.524 ","End":"08:05.820","Text":"Greater than square and less than triangle."},{"Start":"08:05.820 ","End":"08:11.510","Text":"Then I would be able to replace f prime of c by this expression."},{"Start":"08:11.510 ","End":"08:15.600","Text":"Then I will get that the square,"},{"Start":"08:15.600 ","End":"08:17.730","Text":"some number or expression,"},{"Start":"08:17.730 ","End":"08:21.950","Text":"is less than f of b minus f of"},{"Start":"08:21.950 ","End":"08:28.774","Text":"a over b minus a is less than the other expression triangle."},{"Start":"08:28.774 ","End":"08:36.469","Text":"This will typically be of the form that we have to show in the statement of the question."},{"Start":"08:36.469 ","End":"08:43.235","Text":"Sometimes it will be better to multiply out by b minus a so we might get something"},{"Start":"08:43.235 ","End":"08:47.870","Text":"like b minus a times"},{"Start":"08:47.870 ","End":"08:54.215","Text":"something is less than f of b minus f of a,"},{"Start":"08:54.215 ","End":"09:01.400","Text":"is less than b minus a times something else."},{"Start":"09:01.400 ","End":"09:04.950","Text":"Notice that b minus a is positive,"},{"Start":"09:04.950 ","End":"09:10.085","Text":"a is less than b so the inequality is okay."},{"Start":"09:10.085 ","End":"09:15.485","Text":"The confusing part is how to get this initial estimation."},{"Start":"09:15.485 ","End":"09:21.540","Text":"It usually follows from this either"},{"Start":"09:21.540 ","End":"09:27.410","Text":"by some computation or if f is more general, not given explicitly."},{"Start":"09:27.410 ","End":"09:29.135","Text":"The question might state,"},{"Start":"09:29.135 ","End":"09:34.135","Text":"given that f prime c is between 3 and 5."},{"Start":"09:34.135 ","End":"09:38.245","Text":"The examples will explain there\u0027s no point in me hand-waving."},{"Start":"09:38.245 ","End":"09:42.730","Text":"You\u0027ll be absolute wizards of this thing after a few examples."},{"Start":"09:42.730 ","End":"09:46.225","Text":"Without further ado, let\u0027s get onto the first example."},{"Start":"09:46.225 ","End":"09:53.935","Text":"It\u0027s an example exercise and I don\u0027t want to waste your time so I wrote it in advance."},{"Start":"09:53.935 ","End":"09:58.870","Text":"What it ask us to do is to prove the following inequalities."},{"Start":"09:58.870 ","End":"10:01.840","Text":"Here it is. Just take a look at it."},{"Start":"10:01.840 ","End":"10:08.485","Text":"Of course, I have to have because of the square root that both a and b are positive."},{"Start":"10:08.485 ","End":"10:14.260","Text":"For Lagrange\u0027s theorem, I\u0027ll be wanting a less than b."},{"Start":"10:14.260 ","End":"10:19.000","Text":"What I\u0027m going to write at the side is that 0 is less than a,"},{"Start":"10:19.000 ","End":"10:20.815","Text":"which is less than b."},{"Start":"10:20.815 ","End":"10:25.135","Text":"That way these will both be positive and a will be less than b."},{"Start":"10:25.135 ","End":"10:26.860","Text":"How do we get started?"},{"Start":"10:26.860 ","End":"10:28.420","Text":"What function do we take?"},{"Start":"10:28.420 ","End":"10:33.475","Text":"If you look at this and you look at the fb minus f of a,"},{"Start":"10:33.475 ","End":"10:37.660","Text":"looks very much like our function should be the square root function."},{"Start":"10:37.660 ","End":"10:45.610","Text":"Let\u0027s take f of x equals the square root of x."},{"Start":"10:45.610 ","End":"10:52.900","Text":"The first thing to do is make sure it satisfies the conditions of Lagrange\u0027s theorem."},{"Start":"10:52.900 ","End":"11:01.150","Text":"What we need is that for it to be continuous on a b,"},{"Start":"11:01.150 ","End":"11:10.915","Text":"the closed interval and differentiable on the open interval a b."},{"Start":"11:10.915 ","End":"11:12.745","Text":"That\u0027s clearly true."},{"Start":"11:12.745 ","End":"11:14.950","Text":"It\u0027s continuous everywhere."},{"Start":"11:14.950 ","End":"11:21.595","Text":"The square root function is defined for x bigger or equal to 0 in general."},{"Start":"11:21.595 ","End":"11:25.195","Text":"Except for the point 0, it\u0027s differentiable."},{"Start":"11:25.195 ","End":"11:26.905","Text":"That\u0027s no problem."},{"Start":"11:26.905 ","End":"11:30.130","Text":"If it satisfies the condition of the theorem,"},{"Start":"11:30.130 ","End":"11:34.210","Text":"then we can draw the conclusions of the theorem."},{"Start":"11:34.210 ","End":"11:42.310","Text":"The theorem says that there is a point c between a and b."},{"Start":"11:42.310 ","End":"11:44.800","Text":"In other words, there exists,"},{"Start":"11:44.800 ","End":"11:53.500","Text":"such a c that f prime of c is equal to f of b"},{"Start":"11:53.500 ","End":"11:57.970","Text":"minus f of a over b"},{"Start":"11:57.970 ","End":"12:04.030","Text":"minus a. F is not just some abstract function."},{"Start":"12:04.030 ","End":"12:05.560","Text":"We know exactly what it is."},{"Start":"12:05.560 ","End":"12:07.750","Text":"F of x is square root of x."},{"Start":"12:07.750 ","End":"12:11.645","Text":"We can now interpret this and say,"},{"Start":"12:11.645 ","End":"12:13.560","Text":"well, I\u0027ll start with the right-hand side."},{"Start":"12:13.560 ","End":"12:15.015","Text":"It\u0027s just easier."},{"Start":"12:15.015 ","End":"12:19.455","Text":"F of b is the square root of b because f is the square root."},{"Start":"12:19.455 ","End":"12:24.580","Text":"We have minus the square root of a over b minus a."},{"Start":"12:24.580 ","End":"12:26.140","Text":"For the left-hand side,"},{"Start":"12:26.140 ","End":"12:29.125","Text":"if f of x is the square root of x,"},{"Start":"12:29.125 ","End":"12:32.890","Text":"then its derivative f prime is very well known."},{"Start":"12:32.890 ","End":"12:35.830","Text":"It\u0027s 1 over twice the square root of x."},{"Start":"12:35.830 ","End":"12:38.350","Text":"But in our case I\u0027m not going to write x,"},{"Start":"12:38.350 ","End":"12:40.540","Text":"I\u0027m going to write c. Skip the step of"},{"Start":"12:40.540 ","End":"12:43.285","Text":"writing f prime of x is 1 over twice square root of x."},{"Start":"12:43.285 ","End":"12:47.690","Text":"Immediately substituted c. This is what I get."},{"Start":"12:48.420 ","End":"12:51.370","Text":"All I have mainly is equalities."},{"Start":"12:51.370 ","End":"12:54.700","Text":"I want some inequality. I\u0027ll tell you what we\u0027ll do."},{"Start":"12:54.700 ","End":"12:59.200","Text":"We\u0027ll start with this inequality and see if we can gradually change"},{"Start":"12:59.200 ","End":"13:04.135","Text":"c into 1 over twice the square root of c, you\u0027ll see what I mean."},{"Start":"13:04.135 ","End":"13:05.785","Text":"Let\u0027s start from here,"},{"Start":"13:05.785 ","End":"13:09.670","Text":"and I\u0027ll continue over here where I have some space."},{"Start":"13:09.670 ","End":"13:11.755","Text":"Let\u0027s first of all,"},{"Start":"13:11.755 ","End":"13:14.350","Text":"take the square root of everything."},{"Start":"13:14.350 ","End":"13:16.390","Text":"They\u0027re all positive numbers."},{"Start":"13:16.390 ","End":"13:19.075","Text":"If a is less than c is less than b,"},{"Start":"13:19.075 ","End":"13:22.060","Text":"then the square root of a is less than"},{"Start":"13:22.060 ","End":"13:26.485","Text":"the square root of c is less than the square root of b."},{"Start":"13:26.485 ","End":"13:28.000","Text":"If 2 positive numbers,"},{"Start":"13:28.000 ","End":"13:29.050","Text":"1 smaller than the other,"},{"Start":"13:29.050 ","End":"13:31.060","Text":"then so other square roots."},{"Start":"13:31.060 ","End":"13:38.470","Text":"Now it\u0027s square root of c is still not what we have here so let\u0027s multiply by 2 next."},{"Start":"13:38.470 ","End":"13:44.500","Text":"2 is a positive number and we can certainly multiply an inequality by a positive number."},{"Start":"13:44.500 ","End":"13:51.400","Text":"I get 2 root a is less than 2 root c is less than 2 root b."},{"Start":"13:51.400 ","End":"13:58.450","Text":"The next step will be to take the reciprocals 1 over everything,"},{"Start":"13:58.450 ","End":"14:00.040","Text":"because this is what I want."},{"Start":"14:00.040 ","End":"14:06.780","Text":"Here I\u0027ll get 1 over twice the square root of c. I\u0027ll leave the inequality for a moment."},{"Start":"14:06.780 ","End":"14:11.205","Text":"Here. I\u0027ll get 1 over twice the square root of b."},{"Start":"14:11.205 ","End":"14:15.095","Text":"Here I\u0027ll get 1 over twice the square root of a."},{"Start":"14:15.095 ","End":"14:17.650","Text":"Now, what I claim is the following,"},{"Start":"14:17.650 ","End":"14:19.345","Text":"I\u0027ll write it and then I\u0027ll explain it."},{"Start":"14:19.345 ","End":"14:22.600","Text":"I say we have to reverse the direction of the inequalities."},{"Start":"14:22.600 ","End":"14:26.770","Text":"I\u0027m going to put bigger than here and bigger than here."},{"Start":"14:26.770 ","End":"14:32.410","Text":"The reason for this is that a reciprocal reverses an inequality."},{"Start":"14:32.410 ","End":"14:36.400","Text":"Perhaps the easiest is just to give a numerical example."},{"Start":"14:36.400 ","End":"14:45.835","Text":"Here at the side, suppose I have something like 2 is less than 3,"},{"Start":"14:45.835 ","End":"14:49.240","Text":"is less than 4."},{"Start":"14:49.240 ","End":"14:56.220","Text":"Then if I take the reciprocals and I get 1/2, 1/3 and 1/4."},{"Start":"14:56.220 ","End":"14:58.675","Text":"1/2 is bigger than 1/3,"},{"Start":"14:58.675 ","End":"15:00.965","Text":"which is bigger than 1/4."},{"Start":"15:00.965 ","End":"15:04.675","Text":"This numerical example will convince you."},{"Start":"15:04.675 ","End":"15:08.440","Text":"Okay, now we have an inequality and look,"},{"Start":"15:08.440 ","End":"15:11.155","Text":"what I have in the middle of it is this,"},{"Start":"15:11.155 ","End":"15:12.775","Text":"which is just this."},{"Start":"15:12.775 ","End":"15:21.140","Text":"The idea now is to replace 1 over twice square root of c by this."},{"Start":"15:21.210 ","End":"15:28.760","Text":"Also allow me just to change the direction of the inequality."},{"Start":"15:28.830 ","End":"15:32.320","Text":"What I\u0027ll write is 1 over twice square root of c,"},{"Start":"15:32.320 ","End":"15:41.545","Text":"which is the square root of b minus the square root of a over b minus a."},{"Start":"15:41.545 ","End":"15:45.730","Text":"Let me write what\u0027s bigger than over here."},{"Start":"15:45.730 ","End":"15:48.925","Text":"In other words, I\u0027m going to write this over here."},{"Start":"15:48.925 ","End":"15:52.630","Text":"That\u0027s 1 over twice square root of b."},{"Start":"15:52.630 ","End":"15:57.625","Text":"Still dint want an extra line to just write it in reverse order."},{"Start":"15:57.625 ","End":"16:03.230","Text":"Here, 1 over twice the square root of a."},{"Start":"16:04.230 ","End":"16:09.100","Text":"Just like I mentioned earlier,"},{"Start":"16:09.100 ","End":"16:11.890","Text":"often this is what we have to prove and"},{"Start":"16:11.890 ","End":"16:15.640","Text":"often we just have to multiply out by the b minus a,"},{"Start":"16:15.640 ","End":"16:17.335","Text":"which is the case here."},{"Start":"16:17.335 ","End":"16:21.415","Text":"I\u0027m going to multiply everything by b minus a."},{"Start":"16:21.415 ","End":"16:25.390","Text":"But remember that here it is,"},{"Start":"16:25.390 ","End":"16:27.970","Text":"that 0 less than a,"},{"Start":"16:27.970 ","End":"16:31.915","Text":"less than b, which means that b minus a is positive."},{"Start":"16:31.915 ","End":"16:35.935","Text":"It follows from here that b minus a is positive."},{"Start":"16:35.935 ","End":"16:42.249","Text":"I can safely multiply and get that b minus"},{"Start":"16:42.249 ","End":"16:49.465","Text":"a over twice root b is less than multiplying by b minus a just means throwing it out."},{"Start":"16:49.465 ","End":"16:54.160","Text":"Square root of b minus the square root of a, which is less than."},{"Start":"16:54.160 ","End":"17:01.010","Text":"Then I multiply this by b minus a and get b minus a over twice root a."},{"Start":"17:01.530 ","End":"17:04.015","Text":"If you look at this,"},{"Start":"17:04.015 ","End":"17:06.250","Text":"and I don\u0027t know if I can fit everything in,"},{"Start":"17:06.250 ","End":"17:09.700","Text":"but if you look at the exercise that we had to prove,"},{"Start":"17:09.700 ","End":"17:14.605","Text":"it seems to me that this is exactly what we had to prove."},{"Start":"17:14.605 ","End":"17:18.200","Text":"We are done."},{"Start":"17:18.780 ","End":"17:22.345","Text":"We\u0027re done with the example."},{"Start":"17:22.345 ","End":"17:29.080","Text":"I just want to advise you to do a lot of exercises on this subject."},{"Start":"17:29.080 ","End":"17:32.904","Text":"The more exercises you do the easier it will become."},{"Start":"17:32.904 ","End":"17:37.640","Text":"Now I\u0027m done with the clip also. Bye."}],"ID":8639},{"Watched":false,"Name":"Worked example 2","Duration":"5m 36s","ChapterTopicVideoID":8834,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8834.jpeg","UploadDate":"2017-03-06T09:30:59.7370000","DurationForVideoObject":"PT5M36S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.645","Text":"We have to find all the values of c"},{"Start":"00:03.645 ","End":"00:07.470","Text":"which satisfy the conclusion of the mean value theorem,"},{"Start":"00:07.470 ","End":"00:11.730","Text":"Lagrange\u0027s theorem, for the function f of x equals"},{"Start":"00:11.730 ","End":"00:17.235","Text":"this polynomial on the interval from minus 1 to 1."},{"Start":"00:17.235 ","End":"00:22.274","Text":"Polynomials, good for everything continuous differentiable, well behaved."},{"Start":"00:22.274 ","End":"00:33.135","Text":"The f prime of c is equal to f of b minus f of a over b minus a."},{"Start":"00:33.135 ","End":"00:35.610","Text":"Now you notice it says all,"},{"Start":"00:35.610 ","End":"00:38.690","Text":"and remember I told you that the mean value theorem"},{"Start":"00:38.690 ","End":"00:42.410","Text":"guarantees that there exists a value of c,"},{"Start":"00:42.410 ","End":"00:45.080","Text":"but there might be more than 1."},{"Start":"00:45.080 ","End":"00:50.100","Text":"Let\u0027s see, let\u0027s differentiate f"},{"Start":"00:50.100 ","End":"00:55.160","Text":"and make this computation because we have a and b and we have f,"},{"Start":"00:55.160 ","End":"00:58.530","Text":"we have everything, we just don\u0027t have f prime."},{"Start":"00:59.060 ","End":"01:05.855","Text":"I\u0027ll just differentiate it right away and put the value of c in at the same time."},{"Start":"01:05.855 ","End":"01:11.550","Text":"So what I get is that 3x squared,"},{"Start":"01:11.550 ","End":"01:16.365","Text":"so I\u0027m writing 3c squared plus 4x,"},{"Start":"01:16.365 ","End":"01:18.990","Text":"so I\u0027m writing 4c,"},{"Start":"01:18.990 ","End":"01:21.555","Text":"because I\u0027m replacing x with c,"},{"Start":"01:21.555 ","End":"01:26.130","Text":"minus 4, and that\u0027s the left-hand side,"},{"Start":"01:26.130 ","End":"01:29.595","Text":"equals f of b,"},{"Start":"01:29.595 ","End":"01:32.400","Text":"which is, I\u0027ll write this out in full,"},{"Start":"01:32.400 ","End":"01:37.920","Text":"f of 1 minus f of minus"},{"Start":"01:37.920 ","End":"01:46.685","Text":"1 over b minus a is 1 minus minus 1."},{"Start":"01:46.685 ","End":"01:51.920","Text":"So let\u0027s see what we get, 0 minus,"},{"Start":"01:51.920 ","End":"01:57.830","Text":"and this comes out to be if I put minus 1 in here,"},{"Start":"01:57.830 ","End":"02:07.845","Text":"then I\u0027ve got minus 1 plus 2 plus 4 plus 1 is 6."},{"Start":"02:07.845 ","End":"02:11.700","Text":"So it\u0027s 0 minus 6 because the f of a is 6,"},{"Start":"02:11.700 ","End":"02:15.070","Text":"that\u0027s minus, and then over 2."},{"Start":"02:15.070 ","End":"02:22.875","Text":"So minus 6 over 2 is equal to minus 3,"},{"Start":"02:22.875 ","End":"02:26.870","Text":"and now if I bring this to the other side,"},{"Start":"02:26.870 ","End":"02:35.735","Text":"I will get a quadratic equation, 3c squared plus 4c minus 4 plus 3,"},{"Start":"02:35.735 ","End":"02:39.445","Text":"I get minus 1 equals 0."},{"Start":"02:39.445 ","End":"02:43.054","Text":"Let\u0027s do it by the quadratic formula."},{"Start":"02:43.054 ","End":"02:48.980","Text":"So c is got to equal, minus b,"},{"Start":"02:48.980 ","End":"02:55.400","Text":"which is minus 4 plus or minus the square root of b squared minus 4ac."},{"Start":"02:55.400 ","End":"02:57.380","Text":"So b squared is 16,"},{"Start":"02:57.380 ","End":"03:01.040","Text":"minus 4 times 3 times minus 1 is plus 12,"},{"Start":"03:01.040 ","End":"03:10.060","Text":"so it\u0027s 28, all over 2a, which is 6."},{"Start":"03:11.030 ","End":"03:15.089","Text":"Since 28 is 4 times 7,"},{"Start":"03:15.089 ","End":"03:18.960","Text":"this is root 4 times root 7,"},{"Start":"03:18.960 ","End":"03:23.630","Text":"so I can divide everything by 2 and get minus"},{"Start":"03:23.630 ","End":"03:30.590","Text":"2 plus or minus the square root of 7 over 3."},{"Start":"03:30.590 ","End":"03:33.560","Text":"Think about what I said about the square root of 4,"},{"Start":"03:33.560 ","End":"03:36.170","Text":"square root of 7 if you\u0027re not sure about it,"},{"Start":"03:36.170 ","End":"03:39.920","Text":"you\u0027ll soon see that this is okay."},{"Start":"03:39.920 ","End":"03:49.680","Text":"Square root of 7 is approximately equal to 2.65."},{"Start":"03:49.680 ","End":"03:52.560","Text":"We don\u0027t need to exactly."},{"Start":"03:52.560 ","End":"03:54.645","Text":"If that\u0027s the case,"},{"Start":"03:54.645 ","End":"04:01.930","Text":"then I have 2 values of c. Let me take the minus first."},{"Start":"04:01.930 ","End":"04:04.930","Text":"So I\u0027ve got minus 2 minus 2.65,"},{"Start":"04:04.930 ","End":"04:10.930","Text":"minus 4.65 over 3."},{"Start":"04:11.000 ","End":"04:16.739","Text":"If I take the other 1, the plus,"},{"Start":"04:16.739 ","End":"04:22.125","Text":"then I\u0027ve got minus 2 plus 2.65"},{"Start":"04:22.125 ","End":"04:29.170","Text":"over 3 is 0.65 over 3."},{"Start":"04:30.080 ","End":"04:37.300","Text":"What happens is that because we\u0027re on the interval from minus 1 to 1,"},{"Start":"04:37.300 ","End":"04:42.200","Text":"this is obviously less than minus 1 because this thing is below minus 3."},{"Start":"04:42.200 ","End":"04:47.155","Text":"So I can rule this 1 out and I\u0027m just left with this 1,"},{"Start":"04:47.155 ","End":"04:49.010","Text":"which on the calculator,"},{"Start":"04:49.010 ","End":"04:57.065","Text":"if I do, it\u0027s exactly with the minus 2 plus root 7 over 3 comes out,"},{"Start":"04:57.065 ","End":"05:02.790","Text":"I make it 0.215, 3 decimal places,"},{"Start":"05:02.790 ","End":"05:10.930","Text":"it\u0027s still approximate and this will be our answer, 0.215."},{"Start":"05:12.320 ","End":"05:15.750","Text":"Basically, we\u0027re done with this exercise,"},{"Start":"05:15.750 ","End":"05:18.440","Text":"except that I want to point out that don\u0027t assume"},{"Start":"05:18.440 ","End":"05:21.410","Text":"that 1 has to drop out and 1 has to remain."},{"Start":"05:21.410 ","End":"05:25.490","Text":"It just might have been that both values could have been in our interval,"},{"Start":"05:25.490 ","End":"05:28.640","Text":"and in general, there could be more than 1 value of c,"},{"Start":"05:28.640 ","End":"05:32.000","Text":"which satisfies the conclusion of the mean value theorem."},{"Start":"05:32.000 ","End":"05:37.230","Text":"But here, we\u0027re left with this and so that\u0027s done with exercise 2."}],"ID":9093},{"Watched":false,"Name":"worked example 3","Duration":"2m 45s","ChapterTopicVideoID":8835,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8835.jpeg","UploadDate":"2017-03-06T09:31:20.5430000","DurationForVideoObject":"PT2M45S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.650","Text":"But given that f of x is continuous and differentiable on the interval from 6-15."},{"Start":"00:07.650 ","End":"00:10.050","Text":"It\u0027s a closed interval includes the open interval."},{"Start":"00:10.050 ","End":"00:14.160","Text":"So we can apply Lagrange\u0027s mean value theorem."},{"Start":"00:14.160 ","End":"00:17.790","Text":"We\u0027re also told that it satisfies that f of 6,"},{"Start":"00:17.790 ","End":"00:21.825","Text":"that\u0027s f at the left end point is minus 2,"},{"Start":"00:21.825 ","End":"00:27.945","Text":"and that the derivative of f is less than or equal to 10."},{"Start":"00:27.945 ","End":"00:35.235","Text":"It doesn\u0027t say it, but it means at least for all x in the domain of definition."},{"Start":"00:35.235 ","End":"00:37.500","Text":"Any x between 6 and 15 inclusive,"},{"Start":"00:37.500 ","End":"00:39.415","Text":"f-prime is less than or equal to 10."},{"Start":"00:39.415 ","End":"00:44.585","Text":"The question is, what is the largest possible value for f of 15?"},{"Start":"00:44.585 ","End":"00:47.290","Text":"That\u0027s f at the right end point."},{"Start":"00:47.290 ","End":"00:49.970","Text":"Now, it sounds like a difficult question,"},{"Start":"00:49.970 ","End":"00:50.990","Text":"but in actual fact,"},{"Start":"00:50.990 ","End":"00:52.700","Text":"it\u0027s easier than it looks."},{"Start":"00:52.700 ","End":"00:59.830","Text":"Because we already know that a is going to be 6 and b is going to be 15,"},{"Start":"00:59.830 ","End":"01:06.435","Text":"a equals 6, I\u0027m going to put in b equals 15."},{"Start":"01:06.435 ","End":"01:14.490","Text":"I also know that f-prime of c is less than or equal to 10,"},{"Start":"01:14.490 ","End":"01:16.440","Text":"because it\u0027s true for all x."},{"Start":"01:16.440 ","End":"01:24.000","Text":"They even have another bit of information that f of a or f of 6 equal to minus 2."},{"Start":"01:24.000 ","End":"01:26.765","Text":"Now, if we put all that in here,"},{"Start":"01:26.765 ","End":"01:29.900","Text":"let\u0027s work with the right-hand side first."},{"Start":"01:29.900 ","End":"01:34.500","Text":"I get that f of b is"},{"Start":"01:34.500 ","End":"01:42.015","Text":"15 minus f of a and f of a is here."},{"Start":"01:42.015 ","End":"01:50.760","Text":"It\u0027s f of 6 is minus 2/b minus a,"},{"Start":"01:50.760 ","End":"01:55.710","Text":"which is just 15 minus 6,"},{"Start":"01:55.710 ","End":"01:59.040","Text":"is equal to f-prime of C."},{"Start":"01:59.040 ","End":"02:02.525","Text":"But instead of saying equals f-prime of c,"},{"Start":"02:02.525 ","End":"02:06.870","Text":"I can say straight away that this is less than or equal to 10."},{"Start":"02:07.430 ","End":"02:13.440","Text":"Let\u0027s multiply both sides by 9. Why 9?"},{"Start":"02:13.440 ","End":"02:19.320","Text":"Because the denominator here, this is 9."},{"Start":"02:19.320 ","End":"02:22.790","Text":"I accidentally lost a bit of video, never mind."},{"Start":"02:22.790 ","End":"02:23.870","Text":"You can see the end,"},{"Start":"02:23.870 ","End":"02:27.170","Text":"I multiplied both sides by 9,"},{"Start":"02:27.170 ","End":"02:29.415","Text":"which is 15 minus 6."},{"Start":"02:29.415 ","End":"02:32.240","Text":"Then I brought the 2 over to the other side,"},{"Start":"02:32.240 ","End":"02:35.060","Text":"and 90 minus 2 is 88."},{"Start":"02:35.060 ","End":"02:37.190","Text":"That shows us that the largest possible"},{"Start":"02:37.190 ","End":"02:42.795","Text":"value for f of 15 is 88 and that\u0027s the end."},{"Start":"02:42.795 ","End":"02:46.090","Text":"Let\u0027s move on to Exercise 4."}],"ID":9094},{"Watched":false,"Name":"worked example 4","Duration":"2m 2s","ChapterTopicVideoID":8836,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8836.jpeg","UploadDate":"2017-03-06T09:31:35.5330000","DurationForVideoObject":"PT2M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.805","Text":"Here we\u0027re given a function f of x,"},{"Start":"00:02.805 ","End":"00:06.570","Text":"which is continuous and differentiable everywhere."},{"Start":"00:06.570 ","End":"00:09.840","Text":"We\u0027re also told that it has 2 roots,"},{"Start":"00:09.840 ","End":"00:12.555","Text":"could be more, but it has at least 2 roots."},{"Start":"00:12.555 ","End":"00:19.110","Text":"We have to prove that the derivative function f prime of x has at least 1 root."},{"Start":"00:19.110 ","End":"00:24.165","Text":"It sounds like it\u0027s amazingly difficult where actually it\u0027s easy."},{"Start":"00:24.165 ","End":"00:30.345","Text":"You see, all we have to do is to let a and b be the 2 roots,"},{"Start":"00:30.345 ","End":"00:35.040","Text":"and then because of the conditions of continuous and differentiable,"},{"Start":"00:35.040 ","End":"00:42.695","Text":"I can apply Lagrange\u0027s mean value theorem and say that there exists some point c,"},{"Start":"00:42.695 ","End":"00:46.120","Text":"c is between a and b,"},{"Start":"00:46.120 ","End":"00:49.650","Text":"such that f prime of c equals this."},{"Start":"00:49.650 ","End":"00:51.440","Text":"Now, what does it mean to be a root?"},{"Start":"00:51.440 ","End":"00:54.440","Text":"A root of f is where f is 0."},{"Start":"00:54.440 ","End":"00:56.090","Text":"What I get is,"},{"Start":"00:56.090 ","End":"01:03.620","Text":"is that f prime of this c that was guaranteed by Lagrange\u0027s theorem will equal."},{"Start":"01:03.620 ","End":"01:08.960","Text":"Now f of b is 0 because b is a root and a is a root also."},{"Start":"01:08.960 ","End":"01:13.580","Text":"So f of a is also 0 and b is always bigger than a so this"},{"Start":"01:13.580 ","End":"01:19.685","Text":"is not the 0 on the bottom so this expression is just equal to 0."},{"Start":"01:19.685 ","End":"01:25.525","Text":"That means that for x equals c,"},{"Start":"01:25.525 ","End":"01:28.465","Text":"f of x equals 0,"},{"Start":"01:28.465 ","End":"01:32.485","Text":"which means that f has a root."},{"Start":"01:32.485 ","End":"01:37.605","Text":"That root is c. Now it could have more roots,"},{"Start":"01:37.605 ","End":"01:42.085","Text":"but I just know that at least for the c. In general,"},{"Start":"01:42.085 ","End":"01:49.660","Text":"if you have a normal behaved function f and it has 2 roots at least,"},{"Start":"01:49.660 ","End":"01:52.405","Text":"then its derivative has at least 1 root."},{"Start":"01:52.405 ","End":"01:57.775","Text":"That\u0027s quite powerful and very easy with the Lagrange mean value theorem."},{"Start":"01:57.775 ","End":"02:02.930","Text":"We are done with the exercise and the whole clip."}],"ID":9095},{"Watched":false,"Name":"Exercise 1","Duration":"2m 52s","ChapterTopicVideoID":6249,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6249.jpeg","UploadDate":"2017-01-26T05:24:29.1700000","DurationForVideoObject":"PT2M52S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.160","Text":"In this exercise, we have to prove"},{"Start":"00:02.160 ","End":"00:05.985","Text":"this given inequality with a given limitations on A and B."},{"Start":"00:05.985 ","End":"00:08.190","Text":"We\u0027re going to do it with Lagrange\u0027s theorem."},{"Start":"00:08.190 ","End":"00:11.220","Text":"Doesn\u0027t quite look like the setup for Lagrange\u0027s theorem."},{"Start":"00:11.220 ","End":"00:14.580","Text":"But if I rephrase the original question by replacing"},{"Start":"00:14.580 ","End":"00:18.195","Text":"this logarithm of b over a by log b minus log a,"},{"Start":"00:18.195 ","End":"00:22.290","Text":"then it\u0027s a lot more helpful because it really reminds us of Lagrange\u0027s theorem,"},{"Start":"00:22.290 ","End":"00:24.945","Text":"which is based on f of b minus f of a."},{"Start":"00:24.945 ","End":"00:29.000","Text":"So the function to take would be f of x equals natural log of x."},{"Start":"00:29.000 ","End":"00:30.330","Text":"That\u0027s what we\u0027re going to do."},{"Start":"00:30.330 ","End":"00:32.760","Text":"If I let f of x equals natural log of x,"},{"Start":"00:32.760 ","End":"00:35.250","Text":"and notice that it\u0027s continuous and differentiable."},{"Start":"00:35.250 ","End":"00:40.110","Text":"It\u0027s basically everywhere for all positive x and in particular on any closed interval a,"},{"Start":"00:40.110 ","End":"00:43.475","Text":"b, where a and b are both positive when b is bigger than a."},{"Start":"00:43.475 ","End":"00:44.690","Text":"So that\u0027s no problem."},{"Start":"00:44.690 ","End":"00:47.270","Text":"We have met the conditions of Lagrange\u0027s theorem and"},{"Start":"00:47.270 ","End":"00:50.255","Text":"so conclusions of the Lagrange\u0027s theorem tell us that,"},{"Start":"00:50.255 ","End":"00:53.120","Text":"there exists some point c between a and b,"},{"Start":"00:53.120 ","End":"00:56.510","Text":"which satisfies that f of b minus f of a over b"},{"Start":"00:56.510 ","End":"01:00.660","Text":"minus a is f prime at that point c. But I know what the function f is."},{"Start":"01:00.660 ","End":"01:03.060","Text":"The function f of x is natural log of x,"},{"Start":"01:03.060 ","End":"01:04.810","Text":"so I can replace everything here."},{"Start":"01:04.810 ","End":"01:08.330","Text":"What I can say is that the natural log of b minus"},{"Start":"01:08.330 ","End":"01:13.160","Text":"natural log of a over b minus a is f prime of c. Now how do I compute that?"},{"Start":"01:13.160 ","End":"01:15.140","Text":"Well, f of x is natural log of x,"},{"Start":"01:15.140 ","End":"01:19.490","Text":"so f prime of x is 1 over x. I\u0027ll skip the step and straight away say that"},{"Start":"01:19.490 ","End":"01:21.830","Text":"f prime of c is 1 over c. I don\u0027t have to"},{"Start":"01:21.830 ","End":"01:24.875","Text":"write it as x and then change x with c. That will be fine."},{"Start":"01:24.875 ","End":"01:27.110","Text":"Next step is to try and get"},{"Start":"01:27.110 ","End":"01:30.470","Text":"some inequality out of this because we need to prove an inequality."},{"Start":"01:30.470 ","End":"01:35.450","Text":"Now, the only inequality around is this inequality so what I\u0027m going to try and"},{"Start":"01:35.450 ","End":"01:41.000","Text":"do is use this inequality to get an inequality for 1 over c. That\u0027s not a big problem."},{"Start":"01:41.000 ","End":"01:43.535","Text":"All you have to do is take the reciprocal of everything."},{"Start":"01:43.535 ","End":"01:46.234","Text":"But when you take the reciprocal of everything,"},{"Start":"01:46.234 ","End":"01:48.515","Text":"you have to also reverse the order."},{"Start":"01:48.515 ","End":"01:51.890","Text":"You either change the signs of the inequalities or you just reverse the order."},{"Start":"01:51.890 ","End":"01:53.930","Text":"Instead of ACB, we\u0027ve got BCA."},{"Start":"01:53.930 ","End":"01:56.030","Text":"Just to remind you why this is so,"},{"Start":"01:56.030 ","End":"01:58.730","Text":"just take a numerical example that if we"},{"Start":"01:58.730 ","End":"02:02.330","Text":"have let\u0027s say instead of this that 2 is less than 3,"},{"Start":"02:02.330 ","End":"02:09.680","Text":"is less than 4, then certainly we have the 1/4 is less than 1/3 is less than 1/2."},{"Start":"02:09.680 ","End":"02:11.840","Text":"I just changed the order and keep the signs."},{"Start":"02:11.840 ","End":"02:14.600","Text":"Next step is we know what 1 over c is so I\u0027m going to"},{"Start":"02:14.600 ","End":"02:17.630","Text":"replace this 1 over c here by what it was here."},{"Start":"02:17.630 ","End":"02:20.240","Text":"So 1 over b is less than 1 over c,"},{"Start":"02:20.240 ","End":"02:23.330","Text":"which is this thing and I put it here which is less than 1 over a,"},{"Start":"02:23.330 ","End":"02:24.860","Text":"and there\u0027s not much left to do here."},{"Start":"02:24.860 ","End":"02:29.495","Text":"All, that remains really is to multiply out by b minus a."},{"Start":"02:29.495 ","End":"02:33.635","Text":"If I do that, I get b minus a over here and b minus a over here."},{"Start":"02:33.635 ","End":"02:36.440","Text":"But notice I was allowed to do that as you can multiply"},{"Start":"02:36.440 ","End":"02:40.120","Text":"inequalities by positive quantities and b minus a is positive."},{"Start":"02:40.120 ","End":"02:43.430","Text":"B minus a is positive because we were given that b is bigger than a."},{"Start":"02:43.430 ","End":"02:45.680","Text":"Finally, I\u0027ll just change back from"},{"Start":"02:45.680 ","End":"02:48.690","Text":"the subtraction form of the logarithm to the quotient form,"},{"Start":"02:48.690 ","End":"02:51.440","Text":"and I\u0027ve got exactly what we had to prove in the beginning,"},{"Start":"02:51.440 ","End":"02:53.700","Text":"and we are done."}],"ID":6263},{"Watched":false,"Name":"Exercise 2","Duration":"2m 56s","ChapterTopicVideoID":6250,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6250.jpeg","UploadDate":"2017-01-26T05:25:24.2730000","DurationForVideoObject":"PT2M56S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.610","Text":"We have to prove the following inequality for a and b between 0 and Pi over 2."},{"Start":"00:05.610 ","End":"00:10.710","Text":"The reason the Pi over 2 is in here is because the tangent is not defined there."},{"Start":"00:10.710 ","End":"00:15.850","Text":"Tangent in fact is differentiable and continuous from minus Pi over 2 to Pi over 2."},{"Start":"00:15.850 ","End":"00:17.534","Text":"Now, we\u0027re talking about tangent."},{"Start":"00:17.534 ","End":"00:21.530","Text":"Obviously that\u0027s the function we\u0027re going to take for Lagrange\u0027s mean value theorem."},{"Start":"00:21.530 ","End":"00:23.325","Text":"What I\u0027m going to say is the following,"},{"Start":"00:23.325 ","End":"00:28.305","Text":"that we have the function f of x equals tangent x and it\u0027s continuous and differentiable,"},{"Start":"00:28.305 ","End":"00:30.750","Text":"as I said, from minus Pi over 2 to Pi over 2,"},{"Start":"00:30.750 ","End":"00:35.160","Text":"certainly from 0 to Pi over 2 and so in any interval ab like this."},{"Start":"00:35.160 ","End":"00:39.320","Text":"So we satisfied the conditions for a and b between as follows."},{"Start":"00:39.320 ","End":"00:43.295","Text":"Since we\u0027ve satisfied the conditions of Lagrange\u0027s mean value theorem,"},{"Start":"00:43.295 ","End":"00:44.750","Text":"we can get the conclusions."},{"Start":"00:44.750 ","End":"00:48.545","Text":"What it means is that we have some c between a and b,"},{"Start":"00:48.545 ","End":"00:51.500","Text":"such that this is the usual formula,"},{"Start":"00:51.500 ","End":"00:54.620","Text":"f of b minus f of a over b minus a equals f"},{"Start":"00:54.620 ","End":"00:58.205","Text":"prime of c. Now we know what f is in our case is tangent of x."},{"Start":"00:58.205 ","End":"01:02.360","Text":"Remember that the derivative of tangent is 1 over cosine squared."},{"Start":"01:02.360 ","End":"01:06.770","Text":"Immediately here, I could write 1 over cosine squared of c,"},{"Start":"01:06.770 ","End":"01:08.270","Text":"as I said, differentiate this,"},{"Start":"01:08.270 ","End":"01:12.845","Text":"but instead of x put c. Now what we have to do is try and somehow"},{"Start":"01:12.845 ","End":"01:17.855","Text":"bound or estimate this 1 over cosine squared c between an upper and a lower limit."},{"Start":"01:17.855 ","End":"01:19.025","Text":"I need an inequality."},{"Start":"01:19.025 ","End":"01:21.860","Text":"The only inequality that I have so far really is this."},{"Start":"01:21.860 ","End":"01:25.100","Text":"I somehow want to use this inequality and get"},{"Start":"01:25.100 ","End":"01:29.625","Text":"some estimate on 1 over cosine squared c. Let\u0027s do this in stages."},{"Start":"01:29.625 ","End":"01:32.810","Text":"We start off with c being between a and b,"},{"Start":"01:32.810 ","End":"01:38.660","Text":"and then we can move to the cosine on the interval from 0 to Pi over 2,"},{"Start":"01:38.660 ","End":"01:44.540","Text":"the cosine is decreasing function so that if c is between a and b in this direction,"},{"Start":"01:44.540 ","End":"01:48.335","Text":"the cosines reverse the direction of the inequality."},{"Start":"01:48.335 ","End":"01:49.580","Text":"This would be the smallest,"},{"Start":"01:49.580 ","End":"01:51.890","Text":"this is in the middle and this is the largest."},{"Start":"01:51.890 ","End":"01:55.070","Text":"I could draw a quick sketch of that just to show you"},{"Start":"01:55.070 ","End":"01:58.940","Text":"the cosine looks like this from 0 to Pi over 2,"},{"Start":"01:58.940 ","End":"02:01.130","Text":"and it continues in all directions."},{"Start":"02:01.130 ","End":"02:04.520","Text":"But in this interval from 0 to pi over 2,"},{"Start":"02:04.520 ","End":"02:07.880","Text":"it is decreasing, which means that we reversed the order."},{"Start":"02:07.880 ","End":"02:10.910","Text":"Continuing, and the cosine are also positive here."},{"Start":"02:10.910 ","End":"02:12.620","Text":"If I square all of these,"},{"Start":"02:12.620 ","End":"02:16.760","Text":"it will still preserve the inequality because if I have positive numbers in a sequence,"},{"Start":"02:16.760 ","End":"02:19.055","Text":"then so are the squares in the same order,"},{"Start":"02:19.055 ","End":"02:22.100","Text":"and then I can say that the reciprocals,"},{"Start":"02:22.100 ","End":"02:23.975","Text":"I have to reverse directions again."},{"Start":"02:23.975 ","End":"02:26.605","Text":"Remember we gave an example, 2 is less than 3,"},{"Start":"02:26.605 ","End":"02:30.200","Text":"less than 4, but a 1/4 is less than a 1/3, less than a 1/2."},{"Start":"02:30.200 ","End":"02:31.339","Text":"But when you do reciprocal,"},{"Start":"02:31.339 ","End":"02:32.420","Text":"you reverse the order,"},{"Start":"02:32.420 ","End":"02:33.710","Text":"so instead of b, c, a again,"},{"Start":"02:33.710 ","End":"02:35.240","Text":"we have a, c, b,"},{"Start":"02:35.240 ","End":"02:39.020","Text":"and then we can replace the 1 over cosine squared"},{"Start":"02:39.020 ","End":"02:42.860","Text":"c by what it\u0027s equal to from here and all I have to do"},{"Start":"02:42.860 ","End":"02:47.390","Text":"is multiply out by b minus a and note that b minus a is"},{"Start":"02:47.390 ","End":"02:49.700","Text":"a positive number because b is bigger than a and you\u0027re"},{"Start":"02:49.700 ","End":"02:52.640","Text":"allowed to multiply an inequality by a positive number."},{"Start":"02:52.640 ","End":"02:54.905","Text":"This is what we were asked to prove,"},{"Start":"02:54.905 ","End":"02:57.150","Text":"so we are done."}],"ID":6264},{"Watched":false,"Name":"Exercise 3","Duration":"2m 30s","ChapterTopicVideoID":6251,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6251.jpeg","UploadDate":"2017-01-26T05:26:09.1730000","DurationForVideoObject":"PT2M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.340","Text":"Here we have another inequality to prove"},{"Start":"00:02.340 ","End":"00:05.175","Text":"and we\u0027re going to use the mean value theorem."},{"Start":"00:05.175 ","End":"00:09.225","Text":"We have to prove this inequality for any a less than b."},{"Start":"00:09.225 ","End":"00:12.810","Text":"By the look of it, the function we need is e to the minus x,"},{"Start":"00:12.810 ","End":"00:15.800","Text":"and e to the minus x happens to be continuous"},{"Start":"00:15.800 ","End":"00:19.670","Text":"and differentiable everywhere from minus infinity to infinity."},{"Start":"00:19.670 ","End":"00:22.880","Text":"We\u0027ve satisfied the conditions of Lagrange\u0027s Theorem,"},{"Start":"00:22.880 ","End":"00:25.010","Text":"now we can reap the conclusion."},{"Start":"00:25.010 ","End":"00:27.680","Text":"The conclusion is that by this mean value theorem,"},{"Start":"00:27.680 ","End":"00:32.435","Text":"there is some c between a and b such that the usual equation holds,"},{"Start":"00:32.435 ","End":"00:33.500","Text":"and that is this."},{"Start":"00:33.500 ","End":"00:35.360","Text":"But since we have what f is,"},{"Start":"00:35.360 ","End":"00:38.000","Text":"which is e to the minus x, we can substitute."},{"Start":"00:38.000 ","End":"00:42.740","Text":"On the left-hand side we just get e to the minus b and e to the minus a here."},{"Start":"00:42.740 ","End":"00:46.940","Text":"The derivative of e to the minus x is minus e to the minus x."},{"Start":"00:46.940 ","End":"00:49.895","Text":"I\u0027ve replaced the x by c, save a step."},{"Start":"00:49.895 ","End":"00:52.490","Text":"We need to get some inequality,"},{"Start":"00:52.490 ","End":"00:55.895","Text":"and we start off with what we know as an inequality,"},{"Start":"00:55.895 ","End":"00:58.370","Text":"which is that c is between a and b. I would like to"},{"Start":"00:58.370 ","End":"01:01.310","Text":"bound this expression minus e to the minus c,"},{"Start":"01:01.310 ","End":"01:04.190","Text":"also above and below to estimate it."},{"Start":"01:04.190 ","End":"01:06.530","Text":"Let\u0027s start with this and build up to this,"},{"Start":"01:06.530 ","End":"01:08.330","Text":"so c is between a and b,"},{"Start":"01:08.330 ","End":"01:09.740","Text":"a is less than c, less than b."},{"Start":"01:09.740 ","End":"01:11.360","Text":"Now I\u0027m building up to it slowly,"},{"Start":"01:11.360 ","End":"01:12.785","Text":"so I need a minus c,"},{"Start":"01:12.785 ","End":"01:14.690","Text":"so let\u0027s multiply everything by minus,"},{"Start":"01:14.690 ","End":"01:16.640","Text":"but remember multiplying an inequality"},{"Start":"01:16.640 ","End":"01:19.440","Text":"by a negative number means reversing the direction."},{"Start":"01:19.440 ","End":"01:21.110","Text":"Instead of the order a, c, b,"},{"Start":"01:21.110 ","End":"01:22.710","Text":"we have the order b, c, a."},{"Start":"01:22.710 ","End":"01:25.805","Text":"Next thing we want to do is take e to the power of everything,"},{"Start":"01:25.805 ","End":"01:31.145","Text":"but the function e to the power of or the exponential function is an increasing function."},{"Start":"01:31.145 ","End":"01:33.035","Text":"If we have an increasing function,"},{"Start":"01:33.035 ","End":"01:34.700","Text":"it preserves the order,"},{"Start":"01:34.700 ","End":"01:37.055","Text":"so b, c, a is still b, c, a."},{"Start":"01:37.055 ","End":"01:39.950","Text":"So e to this, less than e to this, less than e to this,"},{"Start":"01:39.950 ","End":"01:42.530","Text":"because e to the power of is an increasing function,"},{"Start":"01:42.530 ","End":"01:45.405","Text":"I\u0027m going to sketch it, but it looks something like weeping upwards."},{"Start":"01:45.405 ","End":"01:47.975","Text":"Differentiate e to the x, you\u0027ll get e to the x,"},{"Start":"01:47.975 ","End":"01:50.449","Text":"which is positive, derivative is positive,"},{"Start":"01:50.449 ","End":"01:51.830","Text":"so the function is increasing."},{"Start":"01:51.830 ","End":"01:53.330","Text":"We\u0027re almost where we want to get because"},{"Start":"01:53.330 ","End":"01:55.250","Text":"we want a minus c to the minus c."},{"Start":"01:55.250 ","End":"01:57.529","Text":"So another multiplication by minus,"},{"Start":"01:57.529 ","End":"01:58.910","Text":"and because it\u0027s minus, again,"},{"Start":"01:58.910 ","End":"02:00.000","Text":"we\u0027re reversing the directions,"},{"Start":"02:00.000 ","End":"02:03.005","Text":"instead of a b, c, a we\u0027re having an a, c, b."},{"Start":"02:03.005 ","End":"02:05.270","Text":"Replace this by what it\u0027s equal to,"},{"Start":"02:05.270 ","End":"02:08.055","Text":"which is this, and this is just copied."},{"Start":"02:08.055 ","End":"02:13.260","Text":"Now all I have to do is to multiply by b minus a."},{"Start":"02:13.260 ","End":"02:15.109","Text":"B minus a is positive,"},{"Start":"02:15.109 ","End":"02:16.490","Text":"so we can do that."},{"Start":"02:16.490 ","End":"02:22.730","Text":"I multiply the minus here and here by the b minus a and got a minus b."},{"Start":"02:22.730 ","End":"02:24.770","Text":"If you multiply a difference by a minus,"},{"Start":"02:24.770 ","End":"02:26.075","Text":"you reverse the order,"},{"Start":"02:26.075 ","End":"02:29.360","Text":"and this is the expression that was originally required."},{"Start":"02:29.360 ","End":"02:31.500","Text":"So we are done."}],"ID":6265},{"Watched":false,"Name":"Exercise 4","Duration":"2m 11s","ChapterTopicVideoID":6252,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6252.jpeg","UploadDate":"2017-01-26T05:26:49.8830000","DurationForVideoObject":"PT2M11S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.580","Text":"Here, we have another inequality to prove which"},{"Start":"00:02.580 ","End":"00:05.985","Text":"we\u0027re going to do using Lagrange\u0027s mean value theorem,"},{"Start":"00:05.985 ","End":"00:11.280","Text":"we have this inequality and a and b are positive and a is less than b."},{"Start":"00:11.280 ","End":"00:12.570","Text":"By looking at it,"},{"Start":"00:12.570 ","End":"00:17.130","Text":"you can see that the function that starring here is the arc tangent function."},{"Start":"00:17.130 ","End":"00:21.450","Text":"We\u0027re going to let r, f of x be the arctangent of x,"},{"Start":"00:21.450 ","End":"00:24.920","Text":"and it satisfies the conditions of Lagrange\u0027s theorem"},{"Start":"00:24.920 ","End":"00:29.210","Text":"because it\u0027s differentiable and continuous from minus infinity to infinity."},{"Start":"00:29.210 ","End":"00:31.939","Text":"But in particular, on any interval ab,"},{"Start":"00:31.939 ","End":"00:33.980","Text":"as long as a is less than b,"},{"Start":"00:33.980 ","End":"00:35.510","Text":"left has to be less than the right."},{"Start":"00:35.510 ","End":"00:39.980","Text":"We can draw the conclusions of the mean value theorem and deduce that there are"},{"Start":"00:39.980 ","End":"00:44.749","Text":"some c between a and b such that the usual inequality holds."},{"Start":"00:44.749 ","End":"00:46.700","Text":"Let\u0027s see what this means in our case."},{"Start":"00:46.700 ","End":"00:48.875","Text":"Well, we know that f is the arctangent."},{"Start":"00:48.875 ","End":"00:52.055","Text":"We just need to know what is the derivative of arctangent."},{"Start":"00:52.055 ","End":"00:56.345","Text":"But the derivative of arctangent of x is 1 over 1 plus x squared."},{"Start":"00:56.345 ","End":"00:57.530","Text":"But we straight away,"},{"Start":"00:57.530 ","End":"01:00.395","Text":"instead of x, write c and save a step there,"},{"Start":"01:00.395 ","End":"01:03.110","Text":"so arctangent minus arctangent area would be minus"},{"Start":"01:03.110 ","End":"01:06.500","Text":"a is the derivative of arctangent and substituting c,"},{"Start":"01:06.500 ","End":"01:08.075","Text":"1 over 1 plus c squared."},{"Start":"01:08.075 ","End":"01:10.430","Text":"Now, we haven\u0027t done any inequalities yet."},{"Start":"01:10.430 ","End":"01:12.230","Text":"What we\u0027re going to do is we\u0027re going to start from"},{"Start":"01:12.230 ","End":"01:16.250","Text":"this inequality, and gradually, build up to an inequality involving"},{"Start":"01:16.250 ","End":"01:19.710","Text":"1 over 1 plus c squared instead of c. First thing we\u0027re going"},{"Start":"01:19.710 ","End":"01:23.430","Text":"to do is just copy that c is bigger than A and less than b,"},{"Start":"01:23.430 ","End":"01:27.050","Text":"and then if I square everything because we\u0027re all positive,"},{"Start":"01:27.050 ","End":"01:28.805","Text":"A is bigger than 0,"},{"Start":"01:28.805 ","End":"01:31.520","Text":"the squaring doesn\u0027t change the inequality."},{"Start":"01:31.520 ","End":"01:33.180","Text":"Something is more than another number,"},{"Start":"01:33.180 ","End":"01:36.485","Text":"if they\u0027re both positive than the squares are also in the same order,"},{"Start":"01:36.485 ","End":"01:39.380","Text":"then we have to add 1, which is no problem."},{"Start":"01:39.380 ","End":"01:42.725","Text":"You can add anything positive or negative to an inequality."},{"Start":"01:42.725 ","End":"01:44.840","Text":"But when we take the reciprocal,"},{"Start":"01:44.840 ","End":"01:46.490","Text":"that\u0027s where we have to be careful."},{"Start":"01:46.490 ","End":"01:49.130","Text":"Now, we\u0027ve seen this trick before that when we take the reciprocal,"},{"Start":"01:49.130 ","End":"01:50.780","Text":"we just reverse the order,"},{"Start":"01:50.780 ","End":"01:52.460","Text":"so that instead of a, c, b,"},{"Start":"01:52.460 ","End":"01:54.275","Text":"we\u0027re going with b, c, a."},{"Start":"01:54.275 ","End":"01:57.110","Text":"Now, this 1 over 1 plus c squared is equal to this,"},{"Start":"01:57.110 ","End":"02:00.410","Text":"so we can just replace it and we get this,"},{"Start":"02:00.410 ","End":"02:03.650","Text":"which is getting very close to the final form."},{"Start":"02:03.650 ","End":"02:06.800","Text":"All I have to do is multiply by b minus a, which is positive."},{"Start":"02:06.800 ","End":"02:09.335","Text":"It\u0027s okay and we get this,"},{"Start":"02:09.335 ","End":"02:11.180","Text":"which is what we originally had to prove"},{"Start":"02:11.180 ","End":"02:12.870","Text":"and we\u0027re done."}],"ID":6266},{"Watched":false,"Name":"Exercise 5","Duration":"2m 37s","ChapterTopicVideoID":6253,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6253.jpeg","UploadDate":"2017-01-26T05:27:42.9500000","DurationForVideoObject":"PT2M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"In this exercise, we have to prove the following inequality."},{"Start":"00:03.480 ","End":"00:07.905","Text":"Clearly, the star of the exercise key function here is the arcsin."},{"Start":"00:07.905 ","End":"00:13.575","Text":"I\u0027d like to point out that the arcsin is actually defined from minus 1-1,"},{"Start":"00:13.575 ","End":"00:15.300","Text":"certainly from 0-1,"},{"Start":"00:15.300 ","End":"00:18.960","Text":"and it\u0027s continuous and differentiable from 0-1."},{"Start":"00:18.960 ","End":"00:23.415","Text":"For any interval a, b between zero and 1, and so,"},{"Start":"00:23.415 ","End":"00:26.700","Text":"because we satisfy the conditions of Lagrange\u0027s Mean Value Theorem,"},{"Start":"00:26.700 ","End":"00:32.010","Text":"we also benefit from the conclusions and we know that there is this function f of x."},{"Start":"00:32.010 ","End":"00:34.635","Text":"We have a point c between a and b,"},{"Start":"00:34.635 ","End":"00:38.705","Text":"such that the usual equality holds as I\u0027ve written here,"},{"Start":"00:38.705 ","End":"00:42.800","Text":"what we have to do now is to substitute instead of f the arcsin function,"},{"Start":"00:42.800 ","End":"00:47.105","Text":"and what we have on the right-hand side is the derivative of the arcsin."},{"Start":"00:47.105 ","End":"00:51.170","Text":"Derivative of arcsin x is 1 over the square root of 1 minus x squared."},{"Start":"00:51.170 ","End":"00:54.140","Text":"I\u0027ve just replaced x by c because we don\u0027t want f of x,"},{"Start":"00:54.140 ","End":"00:57.305","Text":"we want f prime of c. We\u0027re up to this point here."},{"Start":"00:57.305 ","End":"00:59.730","Text":"Now we want to start thinking about inequalities."},{"Start":"00:59.730 ","End":"01:02.930","Text":"Well, as before we start with the inequality that\u0027s here,"},{"Start":"01:02.930 ","End":"01:04.420","Text":"and I\u0027ll just copy it over here."},{"Start":"01:04.420 ","End":"01:09.695","Text":"We try and build our way up to an inequality involving this expression here."},{"Start":"01:09.695 ","End":"01:15.620","Text":"The first step will raise everything to the power of 2 and here\u0027s where the 0 comes in,"},{"Start":"01:15.620 ","End":"01:17.000","Text":"that will be greater than 0,"},{"Start":"01:17.000 ","End":"01:19.110","Text":"because if a and b are bigger than 0,"},{"Start":"01:19.110 ","End":"01:23.285","Text":"so is c then we can raise everything squared installed, preserve the inequality."},{"Start":"01:23.285 ","End":"01:25.580","Text":"Now we\u0027re going to do the 1 minus part."},{"Start":"01:25.580 ","End":"01:26.810","Text":"Well, bit by bit let\u0027s first of all,"},{"Start":"01:26.810 ","End":"01:27.965","Text":"just take the negative."},{"Start":"01:27.965 ","End":"01:29.840","Text":"Of course, when we multiply by a negative,"},{"Start":"01:29.840 ","End":"01:31.640","Text":"we need to reverse the inequality."},{"Start":"01:31.640 ","End":"01:35.315","Text":"The order a, c, b is reversed and it becomes b, c, a."},{"Start":"01:35.315 ","End":"01:37.790","Text":"Next step is to add a 1."},{"Start":"01:37.790 ","End":"01:40.460","Text":"Adding a positive or negative number does not change"},{"Start":"01:40.460 ","End":"01:43.190","Text":"the direction of the inequality and we take a square root."},{"Start":"01:43.190 ","End":"01:45.410","Text":"We also keep the direction b,"},{"Start":"01:45.410 ","End":"01:47.120","Text":"c, and a positive number."},{"Start":"01:47.120 ","End":"01:49.040","Text":"The square root is an increasing function."},{"Start":"01:49.040 ","End":"01:50.630","Text":"But when we take the reciprocal,"},{"Start":"01:50.630 ","End":"01:54.320","Text":"that\u0027s when we again have to reverse and I gave the example before,"},{"Start":"01:54.320 ","End":"01:56.945","Text":"that 2 is less than 3, is less than 4,"},{"Start":"01:56.945 ","End":"02:00.185","Text":"but then we have also that 1/4,"},{"Start":"02:00.185 ","End":"02:01.790","Text":"is less than 1/3,"},{"Start":"02:01.790 ","End":"02:04.340","Text":"is less than 1/2 and this numerical example is"},{"Start":"02:04.340 ","End":"02:07.430","Text":"usually enough to convince anyone that we need to reverse the order."},{"Start":"02:07.430 ","End":"02:11.135","Text":"Okay, next step is to replace the above,"},{"Start":"02:11.135 ","End":"02:13.970","Text":"this 1 over square root of 1 minus c squared."},{"Start":"02:13.970 ","End":"02:18.380","Text":"We\u0027ve bounded it and I\u0027m replacing what this is with this here."},{"Start":"02:18.380 ","End":"02:23.015","Text":"That\u0027s where I\u0027ve replaced this by that because above they were equal."},{"Start":"02:23.015 ","End":"02:26.510","Text":"Final step is just to multiply both sides by b minus a,"},{"Start":"02:26.510 ","End":"02:29.300","Text":"which is usually is legitimate because b is bigger than a,"},{"Start":"02:29.300 ","End":"02:32.660","Text":"so b minus a is positive and we can multiply the inequality."},{"Start":"02:32.660 ","End":"02:38.010","Text":"This, if you look at what we were required to prove in the first place. We\u0027re done."}],"ID":6267},{"Watched":false,"Name":"Exercise 6","Duration":"2m 22s","ChapterTopicVideoID":6254,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6254.jpeg","UploadDate":"2017-01-26T05:28:32.7300000","DurationForVideoObject":"PT2M22S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.680","Text":"As I look at the following exercise,"},{"Start":"00:01.680 ","End":"00:06.660","Text":"I realize the star function here clearly is the hyperbolic arc sine,"},{"Start":"00:06.660 ","End":"00:08.550","Text":"and if you haven\u0027t learned about it,"},{"Start":"00:08.550 ","End":"00:10.740","Text":"you might want to skip this exercise,"},{"Start":"00:10.740 ","End":"00:12.570","Text":"but if not, then we\u0027ll continue."},{"Start":"00:12.570 ","End":"00:18.914","Text":"As I say, the main function here is the hyperbolic arc sine or arc sine hyperbolic,"},{"Start":"00:18.914 ","End":"00:23.190","Text":"and it is continuous and differentiable for all positive numbers."},{"Start":"00:23.190 ","End":"00:25.530","Text":"In fact, for all numbers that are all positive numbers,"},{"Start":"00:25.530 ","End":"00:27.930","Text":"and so for any interval from a to b,"},{"Start":"00:27.930 ","End":"00:29.160","Text":"where a and b are positive."},{"Start":"00:29.160 ","End":"00:32.910","Text":"Now, if I apply Lagrange\u0027s mean value theorem to this interval,"},{"Start":"00:32.910 ","End":"00:35.325","Text":"then what I\u0027ll get is that in theory,"},{"Start":"00:35.325 ","End":"00:38.420","Text":"there exists a number c between a and b,"},{"Start":"00:38.420 ","End":"00:41.179","Text":"which satisfies the usual equation,"},{"Start":"00:41.179 ","End":"00:44.510","Text":"which is this, and if we substitute a set of f,"},{"Start":"00:44.510 ","End":"00:47.570","Text":"the function that we know which is the arc sine hyperbolic,"},{"Start":"00:47.570 ","End":"00:49.535","Text":"then what we get is this,"},{"Start":"00:49.535 ","End":"00:50.825","Text":"and I have to explain,"},{"Start":"00:50.825 ","End":"00:54.110","Text":"the left-hand side is just straightforward replacing f by arc sine."},{"Start":"00:54.110 ","End":"00:56.810","Text":"The right-hand side comes from the immediate derivative."},{"Start":"00:56.810 ","End":"00:59.270","Text":"But the derivative of this arc sine hyperbolic of"},{"Start":"00:59.270 ","End":"01:01.835","Text":"x is 1 over the square root of 1 plus x squared."},{"Start":"01:01.835 ","End":"01:04.220","Text":"I didn\u0027t write x squared, I wrote c squared because right"},{"Start":"01:04.220 ","End":"01:06.985","Text":"away I replaced x by c, save a step."},{"Start":"01:06.985 ","End":"01:08.780","Text":"Now we have to start getting some kind of"},{"Start":"01:08.780 ","End":"01:11.480","Text":"inequality because what we\u0027re asked to prove is that"},{"Start":"01:11.480 ","End":"01:16.120","Text":"this expression here is bounded below and above by some other expression."},{"Start":"01:16.120 ","End":"01:17.210","Text":"We have to evaluate,"},{"Start":"01:17.210 ","End":"01:20.400","Text":"estimate this to be below something and above something."},{"Start":"01:20.400 ","End":"01:21.450","Text":"Let\u0027s work with this,"},{"Start":"01:21.450 ","End":"01:24.395","Text":"and we start from the inequality which is here,"},{"Start":"01:24.395 ","End":"01:25.700","Text":"and I copy it over here,"},{"Start":"01:25.700 ","End":"01:28.070","Text":"and I want to start manipulating this until we"},{"Start":"01:28.070 ","End":"01:30.710","Text":"get an inequality where in the middle, we have this."},{"Start":"01:30.710 ","End":"01:32.120","Text":"So let\u0027s do it step-by-step."},{"Start":"01:32.120 ","End":"01:33.860","Text":"First of all, here we have c,"},{"Start":"01:33.860 ","End":"01:35.960","Text":"here we have first of all c squared."},{"Start":"01:35.960 ","End":"01:38.990","Text":"So let\u0027s square everything and we\u0027re working with positive numbers,"},{"Start":"01:38.990 ","End":"01:41.435","Text":"so it\u0027s no problem, the inequality is preserved."},{"Start":"01:41.435 ","End":"01:44.420","Text":"Next thing is adding or subtracting any number to"},{"Start":"01:44.420 ","End":"01:48.035","Text":"an inequality leaves the direction of the inequality unchanged,"},{"Start":"01:48.035 ","End":"01:49.610","Text":"and similarly, the square root of"},{"Start":"01:49.610 ","End":"01:52.760","Text":"positive numbers preserves the direction of the inequality."},{"Start":"01:52.760 ","End":"01:53.780","Text":"In the next step, however,"},{"Start":"01:53.780 ","End":"01:56.060","Text":"we\u0027re going to take the reciprocals and we\u0027ve seen this before,"},{"Start":"01:56.060 ","End":"01:58.700","Text":"reciprocals reversed the order of the inequality."},{"Start":"01:58.700 ","End":"02:00.470","Text":"So if we had here a, here c,"},{"Start":"02:00.470 ","End":"02:03.680","Text":"here b, here it\u0027s the other way around b, c and a."},{"Start":"02:03.680 ","End":"02:07.040","Text":"At this stage, we have notice here and all I have to do is"},{"Start":"02:07.040 ","End":"02:10.925","Text":"replace this expression by the expression up here, which is this."},{"Start":"02:10.925 ","End":"02:12.600","Text":"So we end up getting this,"},{"Start":"02:12.600 ","End":"02:13.775","Text":"and if you look at this,"},{"Start":"02:13.775 ","End":"02:15.965","Text":"and if you look at what we wrote originally,"},{"Start":"02:15.965 ","End":"02:19.115","Text":"we had to prove, then it\u0027s exactly the same thing."},{"Start":"02:19.115 ","End":"02:21.005","Text":"This has been proved over here,"},{"Start":"02:21.005 ","End":"02:23.370","Text":"and so we are done."}],"ID":6268},{"Watched":false,"Name":"Exercise 7","Duration":"2m ","ChapterTopicVideoID":6255,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6255.jpeg","UploadDate":"2017-01-26T05:29:11.6100000","DurationForVideoObject":"PT2M","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.970","Text":"In this exercise, we have to prove this inequality involving the hyperbolic arc tangent."},{"Start":"00:05.970 ","End":"00:09.119","Text":"If you haven\u0027t learned about inverse hyperbolic functions,"},{"Start":"00:09.119 ","End":"00:11.640","Text":"you may or may not want to skip this exercise."},{"Start":"00:11.640 ","End":"00:16.020","Text":"As I said, the hyperbolic arc tangent is what the main function here."},{"Start":"00:16.020 ","End":"00:19.260","Text":"We know that it\u0027s continuous and differentiable from 0"},{"Start":"00:19.260 ","End":"00:22.760","Text":"to 1 all together and in particular in any interval a,"},{"Start":"00:22.760 ","End":"00:25.260","Text":"b, where a and b are between 0 and 1."},{"Start":"00:25.260 ","End":"00:28.170","Text":"If we use a, b in the Lagrange mean value theorem,"},{"Start":"00:28.170 ","End":"00:32.250","Text":"we conclude that there is some value c between a and b,"},{"Start":"00:32.250 ","End":"00:35.700","Text":"which satisfies our usual equality, which is this."},{"Start":"00:35.700 ","End":"00:39.180","Text":"We\u0027ve done this exercise in different forms many times."},{"Start":"00:39.180 ","End":"00:41.540","Text":"What we do next is we replace f by"},{"Start":"00:41.540 ","End":"00:44.929","Text":"the actual function which is the hyperbolic arc tangent."},{"Start":"00:44.929 ","End":"00:48.980","Text":"We also have to know which derivative which is 1 of those immediate derivatives."},{"Start":"00:48.980 ","End":"00:51.800","Text":"But f prime, which is the derivative of arc-tangent"},{"Start":"00:51.800 ","End":"00:54.920","Text":"hyperbolic is 1 over 1 minus x squared."},{"Start":"00:54.920 ","End":"01:00.035","Text":"As usual, I omit the x part and right away plugging in the c. We\u0027re up to this day."},{"Start":"01:00.035 ","End":"01:02.015","Text":"Now we have to start working on inequalities."},{"Start":"01:02.015 ","End":"01:04.310","Text":"We want to find an inequality for this bit."},{"Start":"01:04.310 ","End":"01:07.450","Text":"We start off by finding an inequality for the right-hand side."},{"Start":"01:07.450 ","End":"01:12.050","Text":"We begin with this inequality as always and I copy it again here."},{"Start":"01:12.050 ","End":"01:13.855","Text":"We manipulate this somehow."},{"Start":"01:13.855 ","End":"01:16.550","Text":"Instead of c, we have 1 over 1 minus c squared."},{"Start":"01:16.550 ","End":"01:17.735","Text":"Then we do it in steps."},{"Start":"01:17.735 ","End":"01:20.740","Text":"The first step is to raise everything to the power of 2."},{"Start":"01:20.740 ","End":"01:24.235","Text":"Since everything\u0027s positive, we keep the direction of the inequality."},{"Start":"01:24.235 ","End":"01:27.380","Text":"Next thing you want to do is to make everything minus."},{"Start":"01:27.380 ","End":"01:29.825","Text":"This of course reverses the inequality."},{"Start":"01:29.825 ","End":"01:33.140","Text":"It goes, this is less than this and this this when it\u0027s with a minus."},{"Start":"01:33.140 ","End":"01:35.405","Text":"Adding 1 doesn\u0027t change anything."},{"Start":"01:35.405 ","End":"01:38.690","Text":"But a reciprocal does reverse the order of the inequality."},{"Start":"01:38.690 ","End":"01:40.985","Text":"When we get 1 over them all down again,"},{"Start":"01:40.985 ","End":"01:43.745","Text":"this goes to the beginning and this goes to the end."},{"Start":"01:43.745 ","End":"01:46.205","Text":"This 1 over 1 minus c squared,"},{"Start":"01:46.205 ","End":"01:48.200","Text":"I can replace with this expression,"},{"Start":"01:48.200 ","End":"01:50.390","Text":"and that gives me the following expression."},{"Start":"01:50.390 ","End":"01:53.765","Text":"All that remains to be done is to multiply both sides"},{"Start":"01:53.765 ","End":"01:57.230","Text":"by b minus a and we\u0027ve got this inequality."},{"Start":"01:57.230 ","End":"02:01.890","Text":"This is exactly what we had to prove and we are done."}],"ID":6269},{"Watched":false,"Name":"Exercise 8","Duration":"3m 10s","ChapterTopicVideoID":6256,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6256.jpeg","UploadDate":"2017-01-26T05:30:10.4300000","DurationForVideoObject":"PT3M10S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.330","Text":"In this exercise, we have to prove this inequality."},{"Start":"00:03.330 ","End":"00:04.740","Text":"I forgot to say what n is."},{"Start":"00:04.740 ","End":"00:06.600","Text":"n is any positive integer,"},{"Start":"00:06.600 ","End":"00:07.890","Text":"presumably from 2 on."},{"Start":"00:07.890 ","End":"00:09.765","Text":"If n is 1, it\u0027s not interesting."},{"Start":"00:09.765 ","End":"00:14.175","Text":"Let\u0027s say that n is equal to positive whole number and a and b"},{"Start":"00:14.175 ","End":"00:18.720","Text":"are positive and it looks like we\u0027re going to use Lagrange\u0027s mean value theorem."},{"Start":"00:18.720 ","End":"00:22.350","Text":"The key function, the main function is the nth root of x."},{"Start":"00:22.350 ","End":"00:26.295","Text":"Let\u0027s call f of x the nth root of x in the alternative notation."},{"Start":"00:26.295 ","End":"00:27.690","Text":"In the exponential notation,"},{"Start":"00:27.690 ","End":"00:30.900","Text":"it\u0027s x to the power of 1 over n. This is always"},{"Start":"00:30.900 ","End":"00:34.380","Text":"continuous and differentiable whenever x is positive."},{"Start":"00:34.380 ","End":"00:36.920","Text":"In particular and in the interval a, b,"},{"Start":"00:36.920 ","End":"00:39.890","Text":"where a and b are positive and b is bigger than a."},{"Start":"00:39.890 ","End":"00:41.630","Text":"We\u0027ve satisfied the conditions of"},{"Start":"00:41.630 ","End":"00:45.290","Text":"the mean value theorem so we can reap the benefits of the conclusion."},{"Start":"00:45.290 ","End":"00:49.730","Text":"The conclusion says that there is some value c between a and b."},{"Start":"00:49.730 ","End":"00:51.110","Text":"We don\u0027t know what it is,"},{"Start":"00:51.110 ","End":"00:54.170","Text":"but it exists such that this equation holds."},{"Start":"00:54.170 ","End":"00:56.640","Text":"Now if we replace f of x,"},{"Start":"00:56.640 ","End":"00:58.460","Text":"and I\u0027ll use the 1 over n form."},{"Start":"00:58.460 ","End":"01:00.590","Text":"Then on the left-hand side we get,"},{"Start":"01:00.590 ","End":"01:04.100","Text":"and here I\u0027m continuing to use the nth root notation,"},{"Start":"01:04.100 ","End":"01:05.270","Text":"but on the right-hand side,"},{"Start":"01:05.270 ","End":"01:07.250","Text":"I\u0027m going to use the exponential notation."},{"Start":"01:07.250 ","End":"01:08.520","Text":"So the derivative will be,"},{"Start":"01:08.520 ","End":"01:10.580","Text":"first of all we multiply by the exponent,"},{"Start":"01:10.580 ","End":"01:11.780","Text":"then reduce it by 1."},{"Start":"01:11.780 ","End":"01:15.965","Text":"Anyway, this is what we get with c replacing the x that we would have got."},{"Start":"01:15.965 ","End":"01:18.560","Text":"Because what we need is f prime of c not f prime of x."},{"Start":"01:18.560 ","End":"01:21.230","Text":"So I\u0027ve replaced the x with c. At this point,"},{"Start":"01:21.230 ","End":"01:24.770","Text":"we want to get an inequality involving this expression."},{"Start":"01:24.770 ","End":"01:28.400","Text":"We begin with this inequality a less than c,"},{"Start":"01:28.400 ","End":"01:30.005","Text":"less than b and from this,"},{"Start":"01:30.005 ","End":"01:35.240","Text":"I say that I can deduce this because 1 over n is going to be less than 1."},{"Start":"01:35.240 ","End":"01:37.970","Text":"It\u0027s going to be a 1/2, a 1/3, a 1/4 and so on."},{"Start":"01:37.970 ","End":"01:41.810","Text":"1 over n minus 1 is going to be negative."},{"Start":"01:41.810 ","End":"01:46.400","Text":"It\u0027s going to be a 1/2 minus 1 or a 1/3 minus 1 or a 1/4 minus 1 or a 1/5 minus 1,"},{"Start":"01:46.400 ","End":"01:50.160","Text":"whatever it is, wherever n is on this list this thing is going to be negative."},{"Start":"01:50.160 ","End":"01:52.835","Text":"Now, when you raise something to a negative power,"},{"Start":"01:52.835 ","End":"01:55.775","Text":"then you reverse the order of the inequality."},{"Start":"01:55.775 ","End":"01:57.560","Text":"The a which was smallest,"},{"Start":"01:57.560 ","End":"02:00.410","Text":"when I raise it to the exponent actually becomes the largest."},{"Start":"02:00.410 ","End":"02:07.145","Text":"Then what I do is divide everything by n because what I had was not this but 1 over n."},{"Start":"02:07.145 ","End":"02:09.845","Text":"I just multiply everything by 1 over n."},{"Start":"02:09.845 ","End":"02:13.265","Text":"Now what I have in the middle is what I have over here."},{"Start":"02:13.265 ","End":"02:16.040","Text":"So I can replace it by this expression here."},{"Start":"02:16.040 ","End":"02:18.845","Text":"First and last things I just copy as is,"},{"Start":"02:18.845 ","End":"02:21.395","Text":"and this I replace by what it\u0027s equal to."},{"Start":"02:21.395 ","End":"02:24.000","Text":"Now, I\u0027m going to separate the exponent."},{"Start":"02:24.000 ","End":"02:26.555","Text":"When we have an exponent of a difference or a sum."},{"Start":"02:26.555 ","End":"02:28.190","Text":"If let\u0027s say, look at it as a sum,"},{"Start":"02:28.190 ","End":"02:30.170","Text":"1 over n plus-minus 1,"},{"Start":"02:30.170 ","End":"02:33.545","Text":"so it\u0027s b_1 over n times b_minus 1."},{"Start":"02:33.545 ","End":"02:38.795","Text":"Similarly, a to the power of this is a_1 over n times a_minus 1."},{"Start":"02:38.795 ","End":"02:41.210","Text":"Little bit of rewriting."},{"Start":"02:41.210 ","End":"02:43.910","Text":"b_1 over n, I\u0027m going back to the root notation,"},{"Start":"02:43.910 ","End":"02:45.545","Text":"is the nth root of b."},{"Start":"02:45.545 ","End":"02:47.600","Text":"The b_minus 1 is 1 over b,"},{"Start":"02:47.600 ","End":"02:50.225","Text":"so b joins the n in the denominator."},{"Start":"02:50.225 ","End":"02:53.525","Text":"Similarly over here, I get the nth root of a over na,"},{"Start":"02:53.525 ","End":"02:55.130","Text":"the middle bit stays the same."},{"Start":"02:55.130 ","End":"02:58.160","Text":"Now, if I multiply all sides by b minus a,"},{"Start":"02:58.160 ","End":"02:59.540","Text":"and remember that b is bigger than a,"},{"Start":"02:59.540 ","End":"03:02.135","Text":"so b minus a is positive. So that\u0027s okay."},{"Start":"03:02.135 ","End":"03:04.460","Text":"We end up with this expression."},{"Start":"03:04.460 ","End":"03:06.680","Text":"If I look at what it said in the beginning,"},{"Start":"03:06.680 ","End":"03:11.190","Text":"you\u0027ll see that this is exactly what we had to prove and so we are done."}],"ID":6270},{"Watched":false,"Name":"Exercise 9","Duration":"6m 55s","ChapterTopicVideoID":6257,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6257.jpeg","UploadDate":"2016-05-27T14:42:56.0770000","DurationForVideoObject":"PT6M55S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.295","Text":"Before I even read the exercise,"},{"Start":"00:03.295 ","End":"00:06.655","Text":"I\u0027d like to say that in this exercise,"},{"Start":"00:06.655 ","End":"00:14.785","Text":"I\u0027m going to be introducing a new technique and that\u0027s why I expect you to be alert."},{"Start":"00:14.785 ","End":"00:17.170","Text":"If you\u0027re not feeling alert,"},{"Start":"00:17.170 ","End":"00:21.340","Text":"then you should consider doing this clip another time,"},{"Start":"00:21.340 ","End":"00:23.665","Text":"you need to be focused and attentive."},{"Start":"00:23.665 ","End":"00:28.315","Text":"Having said that, let me begin right away."},{"Start":"00:28.315 ","End":"00:33.860","Text":"Here we have proved this ugly inequality."},{"Start":"00:35.150 ","End":"00:39.790","Text":"Presumably, we\u0027re going to use Lagrange\u0027s theorem, the mean value theorem."},{"Start":"00:39.790 ","End":"00:41.770","Text":"It doesn\u0027t look like it, but if you rewrite"},{"Start":"00:41.770 ","End":"00:44.840","Text":"the log of the quotient as the difference of the logs,"},{"Start":"00:44.840 ","End":"00:47.315","Text":"what I mean is if you write it like this,"},{"Start":"00:47.315 ","End":"00:52.610","Text":"this middle bit is the natural log of the numerator minus natural log of the denominator."},{"Start":"00:52.610 ","End":"00:54.680","Text":"That looks a lot better."},{"Start":"00:54.680 ","End":"01:00.920","Text":"If we take our function of x to be the natural log of x^2 plus 1,"},{"Start":"01:00.920 ","End":"01:06.245","Text":"then it\u0027s easy to see that this function,"},{"Start":"01:06.245 ","End":"01:08.555","Text":"whenever x is bigger than 0,"},{"Start":"01:08.555 ","End":"01:10.700","Text":"is continuous and differentiable."},{"Start":"01:10.700 ","End":"01:12.650","Text":"In particular in any interval a,"},{"Start":"01:12.650 ","End":"01:15.215","Text":"b for which this holds."},{"Start":"01:15.215 ","End":"01:19.610","Text":"Now we can apply the mean value theorem, and say that,"},{"Start":"01:19.610 ","End":"01:27.680","Text":"for such a and b there is a c between a and b for which the usual formula applies."},{"Start":"01:27.680 ","End":"01:32.240","Text":"If we replace f by what it\u0027s equal to,"},{"Start":"01:32.240 ","End":"01:35.135","Text":"which is the natural log of x^2 plus 1,"},{"Start":"01:35.135 ","End":"01:40.010","Text":"then what we get here is that natural log of"},{"Start":"01:40.010 ","End":"01:45.285","Text":"b^2 plus 1 minus natural log of a^2 plus 1 over b minus a is."},{"Start":"01:45.285 ","End":"01:47.590","Text":"Now, what is f prime?"},{"Start":"01:47.590 ","End":"01:50.550","Text":"The f prime of a natural logarithm,"},{"Start":"01:50.550 ","End":"01:54.585","Text":"we put the x^2 plus 1 in the denominator,"},{"Start":"01:54.585 ","End":"01:57.540","Text":"we put the derivative on top."},{"Start":"01:58.030 ","End":"02:00.790","Text":"What I would have had,"},{"Start":"02:00.790 ","End":"02:04.380","Text":"would have been something like"},{"Start":"02:04.380 ","End":"02:12.070","Text":"2x over x^2 plus 1 for the derivative."},{"Start":"02:12.070 ","End":"02:17.800","Text":"But because we\u0027ve replaced f prime of c,"},{"Start":"02:17.800 ","End":"02:22.530","Text":"then I put c, funny here."},{"Start":"02:22.530 ","End":"02:24.125","Text":"Anyway, you get the point,"},{"Start":"02:24.125 ","End":"02:31.225","Text":"I differentiate as if it was c instead of x and this is what we get."},{"Start":"02:31.225 ","End":"02:36.430","Text":"Now, this 2c over c^2 plus 1 is what\u0027s causing all the difficulty."},{"Start":"02:36.430 ","End":"02:38.740","Text":"That\u0027s why we\u0027re going to need a new technique."},{"Start":"02:38.740 ","End":"02:41.590","Text":"That\u0027s why I\u0027ve also decided to label"},{"Start":"02:41.590 ","End":"02:45.270","Text":"this as a function of c. I was going to call it f of c,"},{"Start":"02:45.270 ","End":"02:47.080","Text":"and then I realized that f is already taken,"},{"Start":"02:47.080 ","End":"02:52.490","Text":"so I called it g of c. Usually in previous exercises,"},{"Start":"02:52.490 ","End":"02:55.160","Text":"we\u0027ve started with a less than c, less than b,"},{"Start":"02:55.160 ","End":"02:57.260","Text":"and somehow by a series of steps,"},{"Start":"02:57.260 ","End":"03:00.890","Text":"gotten an inequality involving what was written here."},{"Start":"03:00.890 ","End":"03:05.060","Text":"The difficulty here is that the c is both on the numerator and the denominator,"},{"Start":"03:05.060 ","End":"03:07.130","Text":"and it just doesn\u0027t work this way."},{"Start":"03:07.130 ","End":"03:11.030","Text":"I\u0027m going to have to use another technique."},{"Start":"03:11.030 ","End":"03:15.550","Text":"As before, I\u0027m going to copy first of all this here."},{"Start":"03:15.550 ","End":"03:21.020","Text":"Now I\u0027m not going to just do algebra that converts c into 2c over c^2 plus 1,"},{"Start":"03:21.020 ","End":"03:24.800","Text":"instead I\u0027m going to conjure up a different technique."},{"Start":"03:24.800 ","End":"03:26.555","Text":"What it says is this."},{"Start":"03:26.555 ","End":"03:29.555","Text":"I know that a less than c, less than b."},{"Start":"03:29.555 ","End":"03:34.650","Text":"Now if I knew that g was an increasing function,"},{"Start":"03:34.650 ","End":"03:36.745","Text":"then it preserves the order,"},{"Start":"03:36.745 ","End":"03:38.540","Text":"than if a less than c, less than b,"},{"Start":"03:38.540 ","End":"03:40.250","Text":"then g of a is less than g of c,"},{"Start":"03:40.250 ","End":"03:42.650","Text":"less than g of b. Conversely,"},{"Start":"03:42.650 ","End":"03:45.365","Text":"if I knew that g was always decreasing,"},{"Start":"03:45.365 ","End":"03:46.750","Text":"then I could say,"},{"Start":"03:46.750 ","End":"03:48.930","Text":"that g of a is bigger than g of c,"},{"Start":"03:48.930 ","End":"03:50.865","Text":"is bigger than g of b."},{"Start":"03:50.865 ","End":"03:57.055","Text":"Now our task is to take this g of c in our case which is 2c over c^2 plus 1,"},{"Start":"03:57.055 ","End":"04:01.570","Text":"and see whether it\u0027s increasing or decreasing."},{"Start":"04:01.570 ","End":"04:05.660","Text":"Now, to differentiate it, g prime of c,"},{"Start":"04:05.660 ","End":"04:09.290","Text":"which is just the derivative of this using the quotient rule,"},{"Start":"04:09.290 ","End":"04:12.870","Text":"I\u0027m not going through every detail, we get this."},{"Start":"04:13.970 ","End":"04:17.775","Text":"Just a second, I\u0027ll scroll up there."},{"Start":"04:17.775 ","End":"04:25.770","Text":"Because this thing is always negative,"},{"Start":"04:25.770 ","End":"04:28.780","Text":"now the reason this is negative,"},{"Start":"04:29.800 ","End":"04:32.630","Text":"I made a small mistake here."},{"Start":"04:32.630 ","End":"04:35.540","Text":"I was going to say that c is bigger than 1."},{"Start":"04:35.540 ","End":"04:44.720","Text":"I think I forgot to write at the very beginning that what we have here is not 0, but 1."},{"Start":"04:44.720 ","End":"04:48.085","Text":"Let me just quickly correct that."},{"Start":"04:48.085 ","End":"04:52.515","Text":"Sorry about that. This is a 1 and this is a 1 here."},{"Start":"04:52.515 ","End":"04:55.670","Text":"Now let me get back to where I was before."},{"Start":"04:55.670 ","End":"05:04.570","Text":"The reason this is decreasing is because since a is bigger than 1, so is c,"},{"Start":"05:05.320 ","End":"05:08.150","Text":"because c is bigger than 1,"},{"Start":"05:08.150 ","End":"05:11.000","Text":"then 1 minus c^2 is negative,"},{"Start":"05:11.000 ","End":"05:12.910","Text":"and this is positive and this is positive,"},{"Start":"05:12.910 ","End":"05:22.350","Text":"so the whole thing ends up being negative less than 0."},{"Start":"05:22.350 ","End":"05:26.760","Text":"That\u0027s why the function g of c,"},{"Start":"05:26.760 ","End":"05:32.600","Text":"which is 2c over c^2 plus 1 is decreasing because this derivative is always negative."},{"Start":"05:32.600 ","End":"05:36.200","Text":"Yeah, this is just such a minor point that I forgot to mention"},{"Start":"05:36.200 ","End":"05:40.730","Text":"that we\u0027re talking about the interval where x is bigger than 1,"},{"Start":"05:40.730 ","End":"05:42.790","Text":"everything\u0027s bigger than 1."},{"Start":"05:42.790 ","End":"05:50.180","Text":"Because it\u0027s decreasing, then we are in the second of these 2 cases."},{"Start":"05:50.180 ","End":"05:56.765","Text":"What we have to say is that g of a is bigger than g of c, bigger than g of b."},{"Start":"05:56.765 ","End":"06:01.410","Text":"This translates into the following;"},{"Start":"06:01.410 ","End":"06:06.005","Text":"when I put g of c is what it was before and then g of a,"},{"Start":"06:06.005 ","End":"06:10.230","Text":"it\u0027s just this and g of b is this."},{"Start":"06:12.170 ","End":"06:17.060","Text":"Replacing the middle bit by what it was originally,"},{"Start":"06:17.060 ","End":"06:21.960","Text":"remember that 2c over c^2 plus 1 was equal to this."},{"Start":"06:21.960 ","End":"06:23.700","Text":"I\u0027m just replacing it."},{"Start":"06:23.700 ","End":"06:28.450","Text":"I get this is between somewhere between this and this."},{"Start":"06:28.550 ","End":"06:37.079","Text":"All I have to do now is multiply both sides by b minus a."},{"Start":"06:37.670 ","End":"06:45.035","Text":"Finally, just rewrite this difference of logarithms in the original form,"},{"Start":"06:45.035 ","End":"06:48.080","Text":"which is the log of the quotient."},{"Start":"06:48.080 ","End":"06:55.510","Text":"This is exactly what we were asked to prove and we have proven it. We\u0027re done."}],"ID":6271},{"Watched":false,"Name":"The Mean Value Theorem - a Variation","Duration":"11m 40s","ChapterTopicVideoID":6277,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6277.jpeg","UploadDate":"2019-11-14T07:19:36.5070000","DurationForVideoObject":"PT11M40S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.100","Text":"In this clip I\u0027m going to be continuing with Lagrange\u0027s mean value theorem."},{"Start":"00:05.100 ","End":"00:09.825","Text":"But this time I\u0027m going to introduce a variation of it."},{"Start":"00:09.825 ","End":"00:12.840","Text":"Very similar but slightly different."},{"Start":"00:12.840 ","End":"00:16.170","Text":"I went back to what I wrote before."},{"Start":"00:16.170 ","End":"00:18.180","Text":"This is exactly what I wrote before,"},{"Start":"00:18.180 ","End":"00:20.175","Text":"and I\u0027ll go over it again."},{"Start":"00:20.175 ","End":"00:24.480","Text":"Says that, if a function f satisfies two things,"},{"Start":"00:24.480 ","End":"00:27.510","Text":"that it\u0027s continuous on the interval from a to b,"},{"Start":"00:27.510 ","End":"00:31.125","Text":"the closed interval which means that it includes a and b,"},{"Start":"00:31.125 ","End":"00:36.060","Text":"and it\u0027s differentiable on the open interval from a to b which means that it\u0027s"},{"Start":"00:36.060 ","End":"00:41.300","Text":"differentiable between a and b but I don\u0027t care what happens exactly at a and b."},{"Start":"00:41.300 ","End":"00:45.005","Text":"Then we\u0027re guaranteed that there is some point c between"},{"Start":"00:45.005 ","End":"00:49.520","Text":"a and b which satisfies the following equality."},{"Start":"00:49.520 ","End":"00:52.415","Text":"I guess I should point out that, of course,"},{"Start":"00:52.415 ","End":"00:55.055","Text":"a has to be less than b and all of these otherwise,"},{"Start":"00:55.055 ","End":"00:56.975","Text":"I think makes sense."},{"Start":"00:56.975 ","End":"01:00.665","Text":"Now, I\u0027m going to introduce a slight variation."},{"Start":"01:00.665 ","End":"01:03.545","Text":"It\u0027s still the same Lagrange mean value theorem,"},{"Start":"01:03.545 ","End":"01:06.680","Text":"but from a slightly different viewpoint."},{"Start":"01:06.680 ","End":"01:11.930","Text":"Now I hope you did all the exercises that related to the previous clip."},{"Start":"01:11.930 ","End":"01:16.270","Text":"Notice that in pretty much all the questions there were two variables,"},{"Start":"01:16.270 ","End":"01:19.700","Text":"or at least b and a were considered as variables."},{"Start":"01:19.700 ","End":"01:22.785","Text":"Here\u0027s the thing about parameters."},{"Start":"01:22.785 ","End":"01:25.760","Text":"There\u0027s a standard joke which isn\u0027t really a joke"},{"Start":"01:25.760 ","End":"01:29.435","Text":"that the definition of a parameter is a variable constant."},{"Start":"01:29.435 ","End":"01:33.555","Text":"This is true because on the one hand,"},{"Start":"01:33.555 ","End":"01:35.510","Text":"b and a are always parameters."},{"Start":"01:35.510 ","End":"01:39.590","Text":"They are given numbers but at this level we don\u0027t know what those numbers are."},{"Start":"01:39.590 ","End":"01:45.450","Text":"What I\u0027m proposing is that we only have one variable and that b should be a"},{"Start":"01:45.450 ","End":"01:51.815","Text":"variable and a should be an actual constant like 0 or 2 or something."},{"Start":"01:51.815 ","End":"01:56.270","Text":"In summary, what I\u0027m going to do now is replace b everywhere by"},{"Start":"01:56.270 ","End":"02:00.980","Text":"x and also view a as a constant and not a variable."},{"Start":"02:00.980 ","End":"02:04.010","Text":"In fact, it will often be a constant 0."},{"Start":"02:04.010 ","End":"02:06.155","Text":"Here it goes with the erasing."},{"Start":"02:06.155 ","End":"02:08.035","Text":"I\u0027ll erase b everywhere."},{"Start":"02:08.035 ","End":"02:15.375","Text":"Here, it\u0027s 1, 2, 3 4, 5,"},{"Start":"02:15.375 ","End":"02:25.295","Text":"6, 1, 2, 3, 4, 5, 6."},{"Start":"02:25.295 ","End":"02:32.540","Text":"This is basically the one-variable variation of Lagrange\u0027s mean value theorem."},{"Start":"02:32.540 ","End":"02:35.720","Text":"But I would like to say that in most cases,"},{"Start":"02:35.720 ","End":"02:38.895","Text":"I\u0027m not guaranteeing it but very often,"},{"Start":"02:38.895 ","End":"02:41.265","Text":"a happens to be 0."},{"Start":"02:41.265 ","End":"02:46.920","Text":"I\u0027ll just rewrite this now with a is equal to 0."},{"Start":"02:47.630 ","End":"02:51.650","Text":"Now we have a special case of variation."},{"Start":"02:51.650 ","End":"03:00.890","Text":"That is, if the function f is continuous and differentiable from 0 to x,"},{"Start":"03:00.890 ","End":"03:03.799","Text":"and if x is bigger than 0,"},{"Start":"03:03.799 ","End":"03:10.125","Text":"then there exists c such that c is between 0 and x and the following holds."},{"Start":"03:10.125 ","End":"03:14.815","Text":"I didn\u0027t have to write the minus 0 but it just looks more similar."},{"Start":"03:14.815 ","End":"03:18.785","Text":"This will be over 90 percent of the cases that a will be 0."},{"Start":"03:18.785 ","End":"03:25.220","Text":"If not it\u0027ll be told it\u0027s specifically like minus Pi or 3 or some other number."},{"Start":"03:25.220 ","End":"03:28.540","Text":"Let\u0027s go and do an exercise."},{"Start":"03:28.540 ","End":"03:33.114","Text":"I pre-wrote the example just to save time."},{"Start":"03:33.114 ","End":"03:37.180","Text":"Here is our exercise example to prove"},{"Start":"03:37.180 ","End":"03:39.940","Text":"the following inequality that this is less than"},{"Start":"03:39.940 ","End":"03:43.845","Text":"this and less than this for x bigger than 0."},{"Start":"03:43.845 ","End":"03:46.650","Text":"It looks like the x bigger than 0."},{"Start":"03:46.650 ","End":"03:52.225","Text":"Here it looks like we\u0027re going to have a equals 0 and be left with just the variable x."},{"Start":"03:52.225 ","End":"03:55.300","Text":"It seems clear at least to me that"},{"Start":"03:55.300 ","End":"03:59.290","Text":"the function that we want to take is this function here."},{"Start":"03:59.290 ","End":"04:09.020","Text":"We should take f of x is equal to natural log of 1 plus x in general."},{"Start":"04:09.080 ","End":"04:14.435","Text":"Notice that if I substitute 0,"},{"Start":"04:14.435 ","End":"04:19.680","Text":"f of 0 is natural log of 1 plus 0,"},{"Start":"04:20.030 ","End":"04:25.180","Text":"which is natural log of 1 which is 0."},{"Start":"04:25.790 ","End":"04:29.390","Text":"Now I have to pause here because I was about to"},{"Start":"04:29.390 ","End":"04:34.225","Text":"use Lagrange\u0027s mean value theorem without verifying the conditions."},{"Start":"04:34.225 ","End":"04:39.300","Text":"This is where I want to go back and look at this function which is natural log of"},{"Start":"04:39.300 ","End":"04:46.069","Text":"1 plus x and see if it really does satisfy the conditions on the interval."},{"Start":"04:46.069 ","End":"04:50.510","Text":"In our case, the interval would be from 0 to x."},{"Start":"04:50.510 ","End":"04:53.104","Text":"Let me just write that interval here."},{"Start":"04:53.104 ","End":"04:58.550","Text":"On the interval from 0 to x,"},{"Start":"04:58.550 ","End":"05:02.120","Text":"the closed interval, this is certainly continuous."},{"Start":"05:02.120 ","End":"05:03.500","Text":"It\u0027s defined everywhere."},{"Start":"05:03.500 ","End":"05:07.039","Text":"It\u0027s defined at 0. It\u0027s defined for bigger than 0"},{"Start":"05:07.039 ","End":"05:11.630","Text":"because this argument will be bigger or equal"},{"Start":"05:11.630 ","End":"05:16.670","Text":"to 1 and certainly natural logarithm"},{"Start":"05:16.670 ","End":"05:20.525","Text":"is defined for bigger or equal to 1 and it\u0027s a continuous function."},{"Start":"05:20.525 ","End":"05:25.475","Text":"That\u0027s for the differentiability from 0 to x."},{"Start":"05:25.475 ","End":"05:32.040","Text":"It\u0027s differentiable for 1 plus x bigger than 0 which"},{"Start":"05:32.040 ","End":"05:39.425","Text":"is differentiable from minus 1 onwards and certainly from 0 onwards."},{"Start":"05:39.425 ","End":"05:43.880","Text":"Everything\u0027s fine as far as continuity and differentiability."},{"Start":"05:43.880 ","End":"05:50.045","Text":"Therefore, I can apply the mean value theorem, Lagrange\u0027s theorem,"},{"Start":"05:50.045 ","End":"05:51.490","Text":"however you want to call it,"},{"Start":"05:51.490 ","End":"05:57.660","Text":"and say that there is some c that will satisfy this."},{"Start":"05:57.950 ","End":"06:00.825","Text":"Why don\u0027t I even use that same color?"},{"Start":"06:00.825 ","End":"06:03.900","Text":"I have a point c that is somewhere"},{"Start":"06:03.900 ","End":"06:11.630","Text":"between 0 and x for which f prime of"},{"Start":"06:11.630 ","End":"06:19.379","Text":"c is equal to f of x minus,"},{"Start":"06:19.379 ","End":"06:24.395","Text":"now, f of 0 we also found out here that it\u0027s equal to 0,"},{"Start":"06:24.395 ","End":"06:28.680","Text":"over x minus 0."},{"Start":"06:29.870 ","End":"06:33.120","Text":"I think I will just simplify this, obviously,"},{"Start":"06:33.120 ","End":"06:36.090","Text":"the minus 0\u0027s are not contributing anything,"},{"Start":"06:36.090 ","End":"06:40.650","Text":"so this is just f of x over x."},{"Start":"06:40.650 ","End":"06:46.895","Text":"Now, we don\u0027t know what f prime of c is because we haven\u0027t done any differentiation yet."},{"Start":"06:46.895 ","End":"06:50.000","Text":"But I\u0027m going to do a differentiation and a substitution,"},{"Start":"06:50.000 ","End":"06:51.725","Text":"two steps in one go."},{"Start":"06:51.725 ","End":"06:59.600","Text":"Look, f prime of c is f prime of x when you substitute x equals c. All"},{"Start":"06:59.600 ","End":"07:02.480","Text":"I have to do is differentiate this and substitute"},{"Start":"07:02.480 ","End":"07:07.880","Text":"c. If I differentiate natural log of 1 plus x,"},{"Start":"07:07.880 ","End":"07:15.154","Text":"I get 1 over 1 plus x because that\u0027s the derivative."},{"Start":"07:15.154 ","End":"07:19.040","Text":"But if I substitute c, then I get 1."},{"Start":"07:19.040 ","End":"07:22.705","Text":"I\u0027m just going to erase the x and put c in its place."},{"Start":"07:22.705 ","End":"07:27.410","Text":"I don\u0027t just want to leave it in terms of f because I know what f is,"},{"Start":"07:27.410 ","End":"07:29.150","Text":"f is natural log of 1 plus x,"},{"Start":"07:29.150 ","End":"07:37.215","Text":"so this is equal to the natural log of 1 plus x over x."},{"Start":"07:37.215 ","End":"07:40.740","Text":"Once again, c is between 0 and x."},{"Start":"07:40.740 ","End":"07:42.555","Text":"Let us not forget that."},{"Start":"07:42.555 ","End":"07:47.795","Text":"At this point is where we want to introduce the inequality or estimation."},{"Start":"07:47.795 ","End":"07:53.180","Text":"I want to estimate 1 over 1 plus c. Knowing that c is between 0 and x,"},{"Start":"07:53.180 ","End":"07:55.910","Text":"what is 1 over 1 plus c going to be?"},{"Start":"07:55.910 ","End":"08:00.800","Text":"We did something like that in the first tutorial."},{"Start":"08:00.800 ","End":"08:05.480","Text":"What we do is we modify this gradually until the c becomes whatever we want here,"},{"Start":"08:05.480 ","End":"08:10.280","Text":"which is 1 over 1 plus c. If I continue from here,"},{"Start":"08:10.280 ","End":"08:14.510","Text":"and I think just use a different color for a change."},{"Start":"08:14.510 ","End":"08:17.580","Text":"I have 0 less than c less than x."},{"Start":"08:17.580 ","End":"08:20.510","Text":"Now, I\u0027m going to add 1 to everything."},{"Start":"08:20.510 ","End":"08:25.085","Text":"Adding something to an inequality is always permissible."},{"Start":"08:25.085 ","End":"08:34.620","Text":"I get 1 is less than 1 plus c which is less than 1 plus x."},{"Start":"08:34.620 ","End":"08:37.775","Text":"Now I\u0027m going to take reciprocals of these."},{"Start":"08:37.775 ","End":"08:40.620","Text":"The reciprocal of 1 is 1,"},{"Start":"08:40.620 ","End":"08:44.580","Text":"the reciprocal of 1 plus c is 1 over 1 plus"},{"Start":"08:44.580 ","End":"08:49.775","Text":"c. The reciprocal of 1 plus x is 1 over 1 plus x."},{"Start":"08:49.775 ","End":"08:52.580","Text":"But wait you say, I forgot the inequalities."},{"Start":"08:52.580 ","End":"08:55.864","Text":"There\u0027s a delicate point here and we\u0027ve seen it before,"},{"Start":"08:55.864 ","End":"08:59.675","Text":"is that when we take reciprocals of positive numbers,"},{"Start":"08:59.675 ","End":"09:03.530","Text":"we need to reverse the direction of the inequality."},{"Start":"09:03.530 ","End":"09:08.810","Text":"For those who are not convinced I just give an example of something arithmetical."},{"Start":"09:08.810 ","End":"09:11.420","Text":"For example, in this case,"},{"Start":"09:11.420 ","End":"09:17.495","Text":"I have 1 is less than 2 is less than 3, let\u0027s say."},{"Start":"09:17.495 ","End":"09:19.835","Text":"But if I take the reciprocal,"},{"Start":"09:19.835 ","End":"09:28.445","Text":"then I get that 1 over 1 which is still 1 is bigger than 1/2 which is bigger than 1/3."},{"Start":"09:28.445 ","End":"09:32.030","Text":"Usually a numerical example like this will convince you."},{"Start":"09:32.030 ","End":"09:36.965","Text":"Now, we\u0027ve got 1 over 1 plus c,"},{"Start":"09:36.965 ","End":"09:40.185","Text":"but this is equal to this."},{"Start":"09:40.185 ","End":"09:42.215","Text":"Now I\u0027m going to substitute,"},{"Start":"09:42.215 ","End":"09:46.445","Text":"go back here and I\u0027ll substitute this inequality."},{"Start":"09:46.445 ","End":"09:48.275","Text":"This takes me back here,"},{"Start":"09:48.275 ","End":"09:51.830","Text":"and then I\u0027ll switch to the other color,"},{"Start":"09:51.830 ","End":"09:55.580","Text":"and I\u0027ll put that if this is between this and this,"},{"Start":"09:55.580 ","End":"09:56.990","Text":"and so is this."},{"Start":"09:56.990 ","End":"09:59.825","Text":"But also I could write it the other way round."},{"Start":"09:59.825 ","End":"10:07.470","Text":"If I change the order and I look at it from this direction with the eye,"},{"Start":"10:07.470 ","End":"10:10.490","Text":"I can say this is less than this and less than this."},{"Start":"10:10.490 ","End":"10:20.150","Text":"I want to say that 1 over 1 plus x is less than something which is less than 1."},{"Start":"10:20.150 ","End":"10:23.750","Text":"I\u0027m just reading it from right to left, the inequality."},{"Start":"10:23.750 ","End":"10:25.310","Text":"What we have here,"},{"Start":"10:25.310 ","End":"10:30.305","Text":"this 1 over 1 plus c is what I have here which is just this,"},{"Start":"10:30.305 ","End":"10:35.075","Text":"so it\u0027s natural log of 1 plus x over x."},{"Start":"10:35.075 ","End":"10:38.120","Text":"We had a minus 0 and a minus 0, they disappeared."},{"Start":"10:38.120 ","End":"10:40.670","Text":"Now we\u0027re still working in positive numbers,"},{"Start":"10:40.670 ","End":"10:43.040","Text":"x is bigger than 0, so why don\u0027t we multiply."},{"Start":"10:43.040 ","End":"10:45.150","Text":"I was just going to say both sides,"},{"Start":"10:45.150 ","End":"10:46.320","Text":"it\u0027s in an inequality."},{"Start":"10:46.320 ","End":"10:49.520","Text":"All the bits of the inequality by this positive number x,"},{"Start":"10:49.520 ","End":"10:56.850","Text":"and then we will get x over 1 plus x."},{"Start":"10:57.770 ","End":"11:00.660","Text":"This is less than,"},{"Start":"11:00.660 ","End":"11:07.585","Text":"multiplying this by x I just get the numerator which is natural log of 1 plus x,"},{"Start":"11:07.585 ","End":"11:11.650","Text":"and multiplying this by x gives me x."},{"Start":"11:11.650 ","End":"11:18.530","Text":"If I now highlight this, let\u0027s see."},{"Start":"11:19.130 ","End":"11:25.610","Text":"Isn\u0027t this exactly the same as what we had to prove?"},{"Start":"11:26.840 ","End":"11:31.420","Text":"This is exactly the same as what we had to prove."},{"Start":"11:31.420 ","End":"11:40.850","Text":"We are definitely done with this exercise as they say in Latin, QED."}],"ID":6262},{"Watched":false,"Name":"Exercise 10","Duration":"3m 36s","ChapterTopicVideoID":6258,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6258.jpeg","UploadDate":"2017-01-26T05:31:15.1830000","DurationForVideoObject":"PT3M36S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"In this exercise, we have to prove the following inequality."},{"Start":"00:03.480 ","End":"00:07.905","Text":"As usual, we\u0027ll be using Lagrange\u0027s mean value theorem,"},{"Start":"00:07.905 ","End":"00:11.370","Text":"x is given to be between 0 and Pi over 2,"},{"Start":"00:11.370 ","End":"00:13.545","Text":"that\u0027s between 0 and 90 degrees."},{"Start":"00:13.545 ","End":"00:16.605","Text":"Everything is defined between 0 and 90 degrees."},{"Start":"00:16.605 ","End":"00:19.770","Text":"The tangent goes haywire at 90 degrees and so does"},{"Start":"00:19.770 ","End":"00:23.415","Text":"the cosine in the denominator because cosine of 90 is 0,"},{"Start":"00:23.415 ","End":"00:24.630","Text":"but up to 90 degrees,"},{"Start":"00:24.630 ","End":"00:26.730","Text":"up to Pi over 2, we\u0027re okay."},{"Start":"00:26.730 ","End":"00:29.865","Text":"No problem in the definition the 0 itself,"},{"Start":"00:29.865 ","End":"00:33.254","Text":"we don\u0027t have a problem either with the tangent or the cosine,"},{"Start":"00:33.254 ","End":"00:34.740","Text":"because cosine of 0 is 1,"},{"Start":"00:34.740 ","End":"00:36.030","Text":"there\u0027s no problem with denominator."},{"Start":"00:36.030 ","End":"00:38.400","Text":"If x is anywhere from 0 to Pi over 2,"},{"Start":"00:38.400 ","End":"00:40.155","Text":"then the intervals from 0 to x,"},{"Start":"00:40.155 ","End":"00:46.025","Text":"including the 0, the function tangent of x is continuous and differentiable on it."},{"Start":"00:46.025 ","End":"00:48.890","Text":"The reason I took tangent is it just seems the obvious thing to do."},{"Start":"00:48.890 ","End":"00:50.070","Text":"That\u0027s the 1 in the middle."},{"Start":"00:50.070 ","End":"00:52.880","Text":"Also, I\u0027m mentally thinking that the derivative of"},{"Start":"00:52.880 ","End":"00:56.705","Text":"tangent is 1 over cosine squared so this is the natural choice."},{"Start":"00:56.705 ","End":"00:58.760","Text":"It\u0027s differentiable on 0,"},{"Start":"00:58.760 ","End":"01:01.950","Text":"x closed interval when x satisfies this constraint,"},{"Start":"01:01.950 ","End":"01:04.160","Text":"so we can draw the conclusions of"},{"Start":"01:04.160 ","End":"01:08.000","Text":"the mean value theorem and say that there is a c between 0 and x,"},{"Start":"01:08.000 ","End":"01:10.910","Text":"which satisfies the usual equation that we write."},{"Start":"01:10.910 ","End":"01:14.645","Text":"Now, again proceeding this is recipe style."},{"Start":"01:14.645 ","End":"01:18.485","Text":"We replace f by the function that it is, which is tangent."},{"Start":"01:18.485 ","End":"01:21.995","Text":"Also notice that f prime, and I mentioned it,"},{"Start":"01:21.995 ","End":"01:25.850","Text":"of tangent is 1 over cosine squared."},{"Start":"01:25.850 ","End":"01:30.020","Text":"What we have is that tangent of x minus tangent of 0,"},{"Start":"01:30.020 ","End":"01:32.585","Text":"and here x minus 0, and here\u0027s the derivative,"},{"Start":"01:32.585 ","End":"01:34.160","Text":"1 over cosine squared."},{"Start":"01:34.160 ","End":"01:36.950","Text":"Now, you have to get an inequality out of this somehow."},{"Start":"01:36.950 ","End":"01:40.295","Text":"Recalling that c is between 0 and x,"},{"Start":"01:40.295 ","End":"01:45.830","Text":"I want to somehow get from c to 1 over cosine squared c. Now,"},{"Start":"01:45.830 ","End":"01:50.000","Text":"in the interval from 0 to 90 degrees,"},{"Start":"01:50.000 ","End":"01:52.730","Text":"the cosine is a decreasing function."},{"Start":"01:52.730 ","End":"01:55.130","Text":"Now we\u0027ve said this before that if I draw"},{"Start":"01:55.130 ","End":"01:59.930","Text":"just a very rough sketch of the cosine from 0 to 90 degrees,"},{"Start":"01:59.930 ","End":"02:03.545","Text":"so the cosine at 0 starts off at 1,"},{"Start":"02:03.545 ","End":"02:06.215","Text":"at Pi over 2, it goes down to 0."},{"Start":"02:06.215 ","End":"02:08.495","Text":"This is the part that we\u0027re interested in."},{"Start":"02:08.495 ","End":"02:10.580","Text":"The cosine is decreasing."},{"Start":"02:10.580 ","End":"02:12.739","Text":"If I apply cosine to this,"},{"Start":"02:12.739 ","End":"02:15.260","Text":"then you get the inequality being reversed."},{"Start":"02:15.260 ","End":"02:17.840","Text":"We start off with the x and the c, then the 0,"},{"Start":"02:17.840 ","End":"02:19.565","Text":"because we have a decreasing function,"},{"Start":"02:19.565 ","End":"02:21.020","Text":"we\u0027ve done this thing before."},{"Start":"02:21.020 ","End":"02:23.930","Text":"Now, I\u0027ve still haven\u0027t got to where I want to go"},{"Start":"02:23.930 ","End":"02:27.580","Text":"so the next step is to square everything and the square,"},{"Start":"02:27.580 ","End":"02:33.050","Text":"these are all positive numbers between 0 and Pi over 2 so the inequality is preserved."},{"Start":"02:33.050 ","End":"02:35.345","Text":"But when I take a reciprocal, as you know,"},{"Start":"02:35.345 ","End":"02:37.985","Text":"we reverse the direction of the inequality,"},{"Start":"02:37.985 ","End":"02:39.800","Text":"or you just reverse the order of the terms,"},{"Start":"02:39.800 ","End":"02:42.665","Text":"so the 0 comes first and the c then the x."},{"Start":"02:42.665 ","End":"02:46.250","Text":"Now 1 over cosine squared c is this expression here."},{"Start":"02:46.250 ","End":"02:50.060","Text":"I can just substitute it and the middle term becomes this."},{"Start":"02:50.060 ","End":"02:52.550","Text":"Then if I simplify this,"},{"Start":"02:52.550 ","End":"02:57.995","Text":"because tangent of 0 is 0 and x minus 0 is x,"},{"Start":"02:57.995 ","End":"03:01.039","Text":"the tangent of 0 is 0."},{"Start":"03:01.039 ","End":"03:05.665","Text":"We also have that x minus 0 is just x."},{"Start":"03:05.665 ","End":"03:08.810","Text":"If I multiply both sides by x,"},{"Start":"03:08.810 ","End":"03:12.340","Text":"then I get that tangent x on its own,"},{"Start":"03:12.340 ","End":"03:15.650","Text":"I multiplied the x here and I multiplied the x here,"},{"Start":"03:15.650 ","End":"03:20.225","Text":"that the cosine of 0 is equal to 1."},{"Start":"03:20.225 ","End":"03:22.700","Text":"That when I multiply this by x,"},{"Start":"03:22.700 ","End":"03:25.130","Text":"it\u0027s x over 1, showed you in the picture,"},{"Start":"03:25.130 ","End":"03:28.160","Text":"the 2 important things about the cosine is that at Pi over 2 is 0,"},{"Start":"03:28.160 ","End":"03:32.885","Text":"at 0 it\u0027s 1 and that\u0027s for the tangent we just need to know the tangent of 0 is 1."},{"Start":"03:32.885 ","End":"03:35.030","Text":"We get to this and this is what we have to prove,"},{"Start":"03:35.030 ","End":"03:37.440","Text":"so we are done."}],"ID":6272},{"Watched":false,"Name":"Exercise 11","Duration":"2m 12s","ChapterTopicVideoID":6259,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6259.jpeg","UploadDate":"2017-01-26T05:31:55.0900000","DurationForVideoObject":"PT2M12S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.775","Text":"Here again, we have an inequality to prove"},{"Start":"00:02.775 ","End":"00:06.330","Text":"and we\u0027re going to do it with Lagrange\u0027s mean value theorem."},{"Start":"00:06.330 ","End":"00:09.360","Text":"The function that starts in this exercise is"},{"Start":"00:09.360 ","End":"00:14.355","Text":"the arctangent function so that\u0027s called that f of x."},{"Start":"00:14.355 ","End":"00:20.400","Text":"Actually, the arctangent is continuous and differentiable everywhere but in particular,"},{"Start":"00:20.400 ","End":"00:24.060","Text":"from 0 to x, as long as x is bigger than 0,"},{"Start":"00:24.060 ","End":"00:26.595","Text":"which it is, otherwise the interval doesn\u0027t make sense."},{"Start":"00:26.595 ","End":"00:30.540","Text":"So it satisfies the conditions of Lagrange\u0027s theorem on this interval."},{"Start":"00:30.540 ","End":"00:35.475","Text":"So we can also draw the conclusions that there is a c between 0 and x,"},{"Start":"00:35.475 ","End":"00:38.130","Text":"which satisfies this usual equation."},{"Start":"00:38.130 ","End":"00:40.070","Text":"Now we substitute instead of f,"},{"Start":"00:40.070 ","End":"00:44.630","Text":"the arctangent and so we end up with this just replacing f by"},{"Start":"00:44.630 ","End":"00:49.790","Text":"arctangent and the derivative of the arctangent is 1 over 1 plus x squared."},{"Start":"00:49.790 ","End":"00:51.700","Text":"But since we wanted at the point c,"},{"Start":"00:51.700 ","End":"00:53.420","Text":"instead of 1 over 1 plus x squared,"},{"Start":"00:53.420 ","End":"00:54.905","Text":"we just replace the x with c,"},{"Start":"00:54.905 ","End":"00:56.165","Text":"and this is what we get."},{"Start":"00:56.165 ","End":"00:58.610","Text":"Now we want to start looking at inequalities."},{"Start":"00:58.610 ","End":"01:03.135","Text":"We start off with this inequality where 0 is less than c is less than x."},{"Start":"01:03.135 ","End":"01:07.595","Text":"We want to somehow get from here to an inequality involving this in steps."},{"Start":"01:07.595 ","End":"01:11.240","Text":"The first step is to square everything and since we have positive numbers,"},{"Start":"01:11.240 ","End":"01:13.940","Text":"the direction of the inequality is preserved."},{"Start":"01:13.940 ","End":"01:16.445","Text":"The next thing we want to do is add 1 to everything."},{"Start":"01:16.445 ","End":"01:20.825","Text":"Adding positive or negative number does not change the direction of the inequality."},{"Start":"01:20.825 ","End":"01:23.825","Text":"However, the reciprocal does reverse it."},{"Start":"01:23.825 ","End":"01:25.400","Text":"We\u0027ve done this before."},{"Start":"01:25.400 ","End":"01:29.630","Text":"What we get is that this is between this and this,"},{"Start":"01:29.630 ","End":"01:30.710","Text":"but in the other direction,"},{"Start":"01:30.710 ","End":"01:34.475","Text":"the 1 plus x squared is on this side and the 1 plus 0 squared on this side"},{"Start":"01:34.475 ","End":"01:38.660","Text":"and if I write out what this 1 over 1 plus c squared is,"},{"Start":"01:38.660 ","End":"01:39.905","Text":"is it still on the screen here,"},{"Start":"01:39.905 ","End":"01:43.385","Text":"which is arctangent of x minus arctangent of 0."},{"Start":"01:43.385 ","End":"01:47.850","Text":"The arctangent of 0 is just 0."},{"Start":"01:47.850 ","End":"01:50.720","Text":"Obviously, x minus 0 is just x."},{"Start":"01:50.720 ","End":"01:55.340","Text":"So we get arctangent of x over x but what I want to do is multiply"},{"Start":"01:55.340 ","End":"02:00.755","Text":"everything by x and then put x in the numerator here and x in the numerator here."},{"Start":"02:00.755 ","End":"02:05.120","Text":"So I end up getting that x over this less than this thing is 1,"},{"Start":"02:05.120 ","End":"02:07.820","Text":"so it becomes x and the x disappears in the denominator."},{"Start":"02:07.820 ","End":"02:13.680","Text":"This is what we were required to prove and we have proven it, so we\u0027re done."}],"ID":6273},{"Watched":false,"Name":"Exercise 12","Duration":"2m 27s","ChapterTopicVideoID":6260,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6260.jpeg","UploadDate":"2017-01-26T05:16:09.8130000","DurationForVideoObject":"PT2M27S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.430","Text":"This exercise involves inverse hyperbolic functions,"},{"Start":"00:03.430 ","End":"00:04.920","Text":"so if you haven\u0027t studied those,"},{"Start":"00:04.920 ","End":"00:05.970","Text":"you might want to skip."},{"Start":"00:05.970 ","End":"00:10.815","Text":"Here, we have the hyperbolic arc sine or arc sine hyperbolic,"},{"Start":"00:10.815 ","End":"00:15.870","Text":"and we have to prove that it satisfies this inequality for positive x."},{"Start":"00:15.870 ","End":"00:21.570","Text":"Now, the function arc sine is continuous and differentiable for x bigger than 0,"},{"Start":"00:21.570 ","End":"00:24.765","Text":"and in particular, including the 0."},{"Start":"00:24.765 ","End":"00:31.665","Text":"In particular, it\u0027s continuous and differentiable in any closed interval from 0 to x,"},{"Start":"00:31.665 ","End":"00:34.140","Text":"provided x is bigger than 0."},{"Start":"00:34.140 ","End":"00:38.655","Text":"That means it satisfies the conditions of Lagrange\u0027s Mean Value Theorem."},{"Start":"00:38.655 ","End":"00:40.290","Text":"We draw all the conclusion from"},{"Start":"00:40.290 ","End":"00:44.225","Text":"the Mean Value Theorem that there is some c between 0 and x,"},{"Start":"00:44.225 ","End":"00:47.150","Text":"for which our usual equality holds."},{"Start":"00:47.150 ","End":"00:49.160","Text":"f prime of c satisfies this."},{"Start":"00:49.160 ","End":"00:50.600","Text":"Now, we know what f is,"},{"Start":"00:50.600 ","End":"00:52.850","Text":"it\u0027s the hyperbolic arc sine."},{"Start":"00:52.850 ","End":"00:55.895","Text":"The derivative of this function, if it was x,"},{"Start":"00:55.895 ","End":"00:59.435","Text":"it would be 1 over the square root of 1 plus x squared."},{"Start":"00:59.435 ","End":"01:02.270","Text":"But since we have to substitute c instead of x,"},{"Start":"01:02.270 ","End":"01:05.855","Text":"I write c straight away,1 over the square root of 1 plus c squared."},{"Start":"01:05.855 ","End":"01:09.185","Text":"Here, just replacing f by arc sine wherever it is."},{"Start":"01:09.185 ","End":"01:11.270","Text":"Now, we need to get an inequality,"},{"Start":"01:11.270 ","End":"01:17.540","Text":"so we build on this inequality that 0 is less than c is less than x,"},{"Start":"01:17.540 ","End":"01:21.590","Text":"and so we have that 0 less than c squared,"},{"Start":"01:21.590 ","End":"01:24.214","Text":"less than x squared just by raising"},{"Start":"01:24.214 ","End":"01:27.005","Text":"positive number and it keeps the order of the inequality."},{"Start":"01:27.005 ","End":"01:29.510","Text":"Adding 1 won\u0027t hurt either."},{"Start":"01:29.510 ","End":"01:32.360","Text":"For positive numbers, we can take the square root,"},{"Start":"01:32.360 ","End":"01:36.469","Text":"but when it comes to taking the reciprocal,1 over,"},{"Start":"01:36.469 ","End":"01:39.770","Text":"then we have to reverse the direction."},{"Start":"01:39.770 ","End":"01:42.260","Text":"This 1 comes first, it\u0027s the smallest,"},{"Start":"01:42.260 ","End":"01:45.985","Text":"that\u0027s the second, that\u0027s the third after we take reciprocals."},{"Start":"01:45.985 ","End":"01:50.100","Text":"Now, replacing the 1 over the square root"},{"Start":"01:50.100 ","End":"01:53.580","Text":"of 1 plus c squared is equal to this expression,"},{"Start":"01:53.580 ","End":"01:55.470","Text":"so instead of this, I\u0027ll put this."},{"Start":"01:55.470 ","End":"02:00.530","Text":"This is what we have, and then all that remains to do is to multiply both sides by x."},{"Start":"02:00.530 ","End":"02:02.525","Text":"The denominator here is just x,"},{"Start":"02:02.525 ","End":"02:04.295","Text":"all this is just x."},{"Start":"02:04.295 ","End":"02:07.430","Text":"This thing is 0 because the sine"},{"Start":"02:07.430 ","End":"02:11.390","Text":"hyperbolic of 0 is 0 so the inverse sine hyperbolic is also."},{"Start":"02:11.390 ","End":"02:15.725","Text":"If we just multiply everything by x and 1 over the square root of 1 is 1,"},{"Start":"02:15.725 ","End":"02:22.520","Text":"then we will get that x over this is less than arc sine hyperbolic of x,"},{"Start":"02:22.520 ","End":"02:24.880","Text":"which is less than 1 times x which is x,"},{"Start":"02:24.880 ","End":"02:26.420","Text":"and this is actually what we had to prove."},{"Start":"02:26.420 ","End":"02:28.440","Text":"So we\u0027re done."}],"ID":6274},{"Watched":false,"Name":"Exercise 13","Duration":"3m 4s","ChapterTopicVideoID":6261,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6261.jpeg","UploadDate":"2017-01-26T05:17:04.0100000","DurationForVideoObject":"PT3M4S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"Here we have another inequality to prove involving inverse hyperbolic functions."},{"Start":"00:05.340 ","End":"00:06.930","Text":"If you haven\u0027t studied those,"},{"Start":"00:06.930 ","End":"00:09.150","Text":"then you might not want to stay,"},{"Start":"00:09.150 ","End":"00:10.545","Text":"but you could stay anyway."},{"Start":"00:10.545 ","End":"00:16.050","Text":"Here\u0027s the inequality that the arc tangent hyperbolic of x is between this and this,"},{"Start":"00:16.050 ","End":"00:19.375","Text":"and x is given to be between 0 and 1."},{"Start":"00:19.375 ","End":"00:22.760","Text":"Main reason for that is that the denominator in"},{"Start":"00:22.760 ","End":"00:26.660","Text":"this region is not 0 so this thing makes sense."},{"Start":"00:26.660 ","End":"00:29.000","Text":"In general, we like to,"},{"Start":"00:29.000 ","End":"00:33.200","Text":"for our inequalities to have things bigger than 0 or between 0 and 1,"},{"Start":"00:33.200 ","End":"00:37.310","Text":"arc tangent itself is defined and continuous and differentiable everywhere."},{"Start":"00:37.310 ","End":"00:38.345","Text":"No problem."},{"Start":"00:38.345 ","End":"00:43.565","Text":"If it\u0027s everywhere then in particular it\u0027s going to be on this interval from 0 to x,"},{"Start":"00:43.565 ","End":"00:46.400","Text":"provided that x is between 0 and 1."},{"Start":"00:46.400 ","End":"00:49.310","Text":"We\u0027ll see the reason for this restriction further on."},{"Start":"00:49.310 ","End":"00:51.530","Text":"I can see why it\u0027s going to be less than 1,"},{"Start":"00:51.530 ","End":"00:53.900","Text":"but the bigger than 0, you\u0027ll see."},{"Start":"00:53.900 ","End":"00:57.800","Text":"We\u0027ve satisfied the conditions of Lagrange\u0027s mean value theorem,"},{"Start":"00:57.800 ","End":"01:02.120","Text":"so we get the conclusions that there is some c between 0 and x,"},{"Start":"01:02.120 ","End":"01:03.940","Text":"which gives us our usual equality,"},{"Start":"01:03.940 ","End":"01:07.445","Text":"f of x minus f of 0 over x minus 0 is f prime"},{"Start":"01:07.445 ","End":"01:11.390","Text":"of c. Now we just substitute for the function,"},{"Start":"01:11.390 ","End":"01:13.130","Text":"the arc tangent hyperbolic."},{"Start":"01:13.130 ","End":"01:16.430","Text":"It\u0027s an immediate derivative that the derivative of"},{"Start":"01:16.430 ","End":"01:20.015","Text":"this function is 1 over 1 minus x squared."},{"Start":"01:20.015 ","End":"01:23.105","Text":"But I write it already with c substituted in."},{"Start":"01:23.105 ","End":"01:24.710","Text":"I don\u0027t write 1 over 1 minus x squared,"},{"Start":"01:24.710 ","End":"01:26.870","Text":"I write it with 1 over 1 minus c squared."},{"Start":"01:26.870 ","End":"01:29.930","Text":"Here I just replace f by arc tangent hyperbolic."},{"Start":"01:29.930 ","End":"01:34.760","Text":"Now, we want to get an inequality out of this somehow and we do it the usual"},{"Start":"01:34.760 ","End":"01:40.280","Text":"way that we begin with this inequality which has to copy over again."},{"Start":"01:40.280 ","End":"01:42.235","Text":"I want to build up to this."},{"Start":"01:42.235 ","End":"01:44.510","Text":"We begin by squaring,"},{"Start":"01:44.510 ","End":"01:46.790","Text":"and that certainly preserves the inequality."},{"Start":"01:46.790 ","End":"01:49.550","Text":"We see the fact that we\u0027re bigger than 0 is"},{"Start":"01:49.550 ","End":"01:52.550","Text":"used here because otherwise you wouldn\u0027t be able to conclude this."},{"Start":"01:52.550 ","End":"01:56.750","Text":"Next, we want to subtract from 1 or multiply by minus and then add 1,"},{"Start":"01:56.750 ","End":"01:58.025","Text":"do it in 2 steps."},{"Start":"01:58.025 ","End":"02:02.030","Text":"Multiply by minus 1 and reverse the order of the terms or you"},{"Start":"02:02.030 ","End":"02:06.155","Text":"could have reversed the arrows and then add 1."},{"Start":"02:06.155 ","End":"02:11.780","Text":"You can always add 1 to any inequality or a positive or negative numbers you can add."},{"Start":"02:11.780 ","End":"02:14.210","Text":"Then we take the reciprocal,"},{"Start":"02:14.210 ","End":"02:17.180","Text":"but reciprocals we know you have to reverse,"},{"Start":"02:17.180 ","End":"02:19.490","Text":"either reverse the arrows, the less than,"},{"Start":"02:19.490 ","End":"02:22.760","Text":"greater than or reverse the order of the terms what we\u0027ve done here."},{"Start":"02:22.760 ","End":"02:24.560","Text":"So 1 over this, less than 1 over this,"},{"Start":"02:24.560 ","End":"02:25.715","Text":"less than 1 over this."},{"Start":"02:25.715 ","End":"02:26.960","Text":"Now this middle thing,"},{"Start":"02:26.960 ","End":"02:29.210","Text":"this 1 over 1 minus c squared,"},{"Start":"02:29.210 ","End":"02:32.270","Text":"if I scroll back up, we see that it\u0027s equal to this expression."},{"Start":"02:32.270 ","End":"02:35.480","Text":"I\u0027m going to put this expression instead of this down here."},{"Start":"02:35.480 ","End":"02:43.475","Text":"We just have to tidy up now because arc tangent hyperbolic of 0 is 0 and this thing here,"},{"Start":"02:43.475 ","End":"02:45.559","Text":"x minus 0 is just x."},{"Start":"02:45.559 ","End":"02:49.235","Text":"If I multiply both sides by x,"},{"Start":"02:49.235 ","End":"02:55.940","Text":"then what I get simply is that this thing is just arc tangent hyperbolic of x."},{"Start":"02:55.940 ","End":"02:58.820","Text":"Here the x goes in the numerator and here 1 over 1,"},{"Start":"02:58.820 ","End":"03:00.560","Text":"which is 1 is multiplied by x."},{"Start":"03:00.560 ","End":"03:05.460","Text":"This is what we get and this is what we were required to prove. We are done."}],"ID":6275},{"Watched":false,"Name":"Exercise 14","Duration":"2m 38s","ChapterTopicVideoID":6262,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6262.jpeg","UploadDate":"2016-05-27T14:44:58.8930000","DurationForVideoObject":"PT2M38S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:03.840","Text":"Here we have to prove this inequality,"},{"Start":"00:03.840 ","End":"00:07.560","Text":"and we\u0027re going to use Lagrange\u0027s mean value theorem;"},{"Start":"00:07.560 ","End":"00:11.490","Text":"the variation form with 0 and x, not the 1 with a and b."},{"Start":"00:11.490 ","End":"00:15.300","Text":"The function to take is clearly e^x."},{"Start":"00:15.300 ","End":"00:20.530","Text":"What we get is that it\u0027s continuous and differentiable."},{"Start":"00:20.570 ","End":"00:23.490","Text":"I\u0027ll just say something about that."},{"Start":"00:23.490 ","End":"00:26.024","Text":"Each of the power of x is continuous and differentiable"},{"Start":"00:26.024 ","End":"00:29.880","Text":"everywhere on the whole number line from minus infinity to infinity."},{"Start":"00:29.880 ","End":"00:32.790","Text":"In particular, it\u0027s going to be differentiable and"},{"Start":"00:32.790 ","End":"00:36.705","Text":"continuous in any particular interval 0 to x,"},{"Start":"00:36.705 ","End":"00:39.225","Text":"as long as x is bigger than 0."},{"Start":"00:39.225 ","End":"00:42.635","Text":"That\u0027s the conditions of the mean value theorem."},{"Start":"00:42.635 ","End":"00:46.370","Text":"The conclusion says that there exists a c between 0 and x,"},{"Start":"00:46.370 ","End":"00:49.370","Text":"which satisfies this condition."},{"Start":"00:49.370 ","End":"00:52.705","Text":"Since f of x is e^x,"},{"Start":"00:52.705 ","End":"00:54.230","Text":"and as a matter of fact,"},{"Start":"00:54.230 ","End":"00:56.645","Text":"f prime of x is also e^x,"},{"Start":"00:56.645 ","End":"00:59.495","Text":"so substituting here is going to be easy."},{"Start":"00:59.495 ","End":"01:04.100","Text":"Just replace everything where f or f prime by e to the power of,"},{"Start":"01:04.100 ","End":"01:08.510","Text":"so this is what we get and just simplifying it slightly."},{"Start":"01:08.510 ","End":"01:10.730","Text":"The e^0 is 1,"},{"Start":"01:10.730 ","End":"01:13.700","Text":"and x minus 0 is x obviously."},{"Start":"01:13.700 ","End":"01:18.930","Text":"What I want to do is get an inequality out of this."},{"Start":"01:21.500 ","End":"01:25.490","Text":"First of all, let\u0027s work with the inequality we had,"},{"Start":"01:25.490 ","End":"01:28.460","Text":"which is this, and copy it over here."},{"Start":"01:28.460 ","End":"01:33.095","Text":"Now we want to see what we can do if we can replace c by e^c."},{"Start":"01:33.095 ","End":"01:36.655","Text":"Well, the exponential function is always increasing."},{"Start":"01:36.655 ","End":"01:37.860","Text":"You draw the graph of it,"},{"Start":"01:37.860 ","End":"01:40.265","Text":"it\u0027ll go something like this."},{"Start":"01:40.265 ","End":"01:44.300","Text":"Therefore, if we take e to the power of everything,"},{"Start":"01:44.300 ","End":"01:47.315","Text":"it\u0027ll preserve the direction of the inequality,"},{"Start":"01:47.315 ","End":"01:49.950","Text":"and this is what we get."},{"Start":"01:51.820 ","End":"02:00.150","Text":"E^0 is 1, and then I replace e^c by what\u0027s it\u0027s equal to,"},{"Start":"02:00.150 ","End":"02:03.670","Text":"which is e^x minus 1 over x."},{"Start":"02:07.640 ","End":"02:11.490","Text":"Next step is to multiply everything by x."},{"Start":"02:11.490 ","End":"02:16.245","Text":"We\u0027ve got x here is less than e^x minus 1,"},{"Start":"02:16.245 ","End":"02:21.710","Text":"which is less than x, e^x."},{"Start":"02:21.710 ","End":"02:26.380","Text":"The last thing I want to do here is just add 1 to everything."},{"Start":"02:26.380 ","End":"02:28.810","Text":"I get here x plus 1."},{"Start":"02:28.810 ","End":"02:30.800","Text":"Here the 1 disappears,"},{"Start":"02:30.800 ","End":"02:33.475","Text":"and here I have 1 plus x e^x,"},{"Start":"02:33.475 ","End":"02:35.960","Text":"and in fact, this is what we had to prove,"},{"Start":"02:35.960 ","End":"02:38.360","Text":"so we are done."}],"ID":6276},{"Watched":false,"Name":"Exercise 15","Duration":"1m 20s","ChapterTopicVideoID":6263,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6263.jpeg","UploadDate":"2017-01-26T05:17:28.2100000","DurationForVideoObject":"PT1M20S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.940","Text":"In this exercise, we\u0027re going to prove the famous inequality that for all positive x,"},{"Start":"00:05.940 ","End":"00:09.239","Text":"the sine of x is less than or equal to x."},{"Start":"00:09.239 ","End":"00:11.490","Text":"We\u0027re going to use, of course,"},{"Start":"00:11.490 ","End":"00:14.085","Text":"Lagrange\u0027s mean value theorem."},{"Start":"00:14.085 ","End":"00:18.810","Text":"The function to take would be the function sine x."},{"Start":"00:18.810 ","End":"00:22.620","Text":"Sine x is continuous and differentiable everywhere."},{"Start":"00:22.620 ","End":"00:25.350","Text":"In particular, for an interval 0,"},{"Start":"00:25.350 ","End":"00:27.450","Text":"x provided that x is bigger than"},{"Start":"00:27.450 ","End":"00:30.885","Text":"0 because the right has to be bigger than the left, of course."},{"Start":"00:30.885 ","End":"00:33.810","Text":"By the conclusions of the mean value theorem,"},{"Start":"00:33.810 ","End":"00:36.390","Text":"the sum c between 0 and x,"},{"Start":"00:36.390 ","End":"00:41.940","Text":"which satisfies the usual equation that you\u0027ve seen dozen times already, just this 1."},{"Start":"00:41.940 ","End":"00:47.370","Text":"Now, we\u0027re going to replace f by the sine and f prime by cosine,"},{"Start":"00:47.370 ","End":"00:52.580","Text":"and we get that sine x minus sine 0 of x minus 0 is the cosine of"},{"Start":"00:52.580 ","End":"00:57.740","Text":"c. Cosine of c is always less than or equal to 1."},{"Start":"00:57.740 ","End":"01:03.445","Text":"Just to remind you that cosine oscillates and it\u0027s always between 1 and minus 1 in fact."},{"Start":"01:03.445 ","End":"01:06.720","Text":"This thing is less than or equal to 1,"},{"Start":"01:06.720 ","End":"01:08.489","Text":"sine 0 is 0,"},{"Start":"01:08.489 ","End":"01:10.575","Text":"so we\u0027ll have the sine x over x,"},{"Start":"01:10.575 ","End":"01:11.820","Text":"less than or equal to 1."},{"Start":"01:11.820 ","End":"01:14.720","Text":"We\u0027re just multiplying by x, which is positive,"},{"Start":"01:14.720 ","End":"01:17.690","Text":"we get that sine x is less than or equal to x,"},{"Start":"01:17.690 ","End":"01:19.265","Text":"which is what was required,"},{"Start":"01:19.265 ","End":"01:21.450","Text":"and we are done."}],"ID":6277},{"Watched":false,"Name":"Exercise 16","Duration":"2m 27s","ChapterTopicVideoID":6264,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6264.jpeg","UploadDate":"2016-05-27T14:45:27.8100000","DurationForVideoObject":"PT2M27S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"In this exercise, we have to prove"},{"Start":"00:02.280 ","End":"00:06.000","Text":"the following inequality that the tangent of x is less than 4x,"},{"Start":"00:06.000 ","End":"00:11.175","Text":"true for x between 0 and this would be 60 degrees."},{"Start":"00:11.175 ","End":"00:15.150","Text":"The function to take would be the tangent of x."},{"Start":"00:15.150 ","End":"00:19.335","Text":"We\u0027ll let f of x be tangent of x."},{"Start":"00:19.335 ","End":"00:27.090","Text":"It\u0027s continuous and differentiable actually between minus Pi over 2 and Pi over 2."},{"Start":"00:27.090 ","End":"00:32.115","Text":"But certainly in this interval between 0 and Pi over 3."},{"Start":"00:32.115 ","End":"00:35.180","Text":"For any sub-interval from 0 to x,"},{"Start":"00:35.180 ","End":"00:38.150","Text":"we satisfied the conditions of Lagrange\u0027s theorem."},{"Start":"00:38.150 ","End":"00:41.645","Text":"We can draw the conclusion that there is some c between"},{"Start":"00:41.645 ","End":"00:47.974","Text":"0 and x such that the usual equality holds."},{"Start":"00:47.974 ","End":"00:51.140","Text":"To develop this, we just replace f by the tangent."},{"Start":"00:51.140 ","End":"00:56.030","Text":"Remember that the derivative of tangent is 1 over cosine squared."},{"Start":"00:56.030 ","End":"01:00.770","Text":"In our case, we have 1 over cosine squared c. I copied"},{"Start":"01:00.770 ","End":"01:05.780","Text":"this inequality and of course combined it with x less than Pi over 3."},{"Start":"01:05.780 ","End":"01:09.390","Text":"This is our present situation."},{"Start":"01:11.590 ","End":"01:16.400","Text":"That means that c is less than Pi over 3 for 1 thing."},{"Start":"01:16.400 ","End":"01:18.110","Text":"I want to build up, instead of c,"},{"Start":"01:18.110 ","End":"01:21.975","Text":"I want to have 1 over cosine squared c and want to have some inequality with it."},{"Start":"01:21.975 ","End":"01:24.635","Text":"If I take, first of all, the cosine,"},{"Start":"01:24.635 ","End":"01:28.265","Text":"the cosine is decreasing from 0 to Pi over 2."},{"Start":"01:28.265 ","End":"01:30.005","Text":"After I take the cosine,"},{"Start":"01:30.005 ","End":"01:33.410","Text":"I have to reverse the terms."},{"Start":"01:33.410 ","End":"01:37.185","Text":"But cosine Pi over 3 is cosine 60 is a 1/2."},{"Start":"01:37.185 ","End":"01:44.645","Text":"I get that 1/2 is less than cosine c. Now, if I square things,"},{"Start":"01:44.645 ","End":"01:48.530","Text":"then I get 1 quarter less than cosine squared c,"},{"Start":"01:48.530 ","End":"01:50.285","Text":"I still need to do a reciprocal."},{"Start":"01:50.285 ","End":"01:56.420","Text":"The reciprocal will reverse the direction of the inequality or the order of the terms."},{"Start":"01:56.420 ","End":"02:03.850","Text":"I get that 1 over cosine squared c is less than reciprocal of a quarter is just 4 again."},{"Start":"02:05.450 ","End":"02:09.200","Text":"1 over cosine squared is just equal to,"},{"Start":"02:09.200 ","End":"02:12.245","Text":"as we saw above, to this expression."},{"Start":"02:12.245 ","End":"02:18.530","Text":"Since tangent of 0 is 0 and this is just tangent x/x,"},{"Start":"02:18.530 ","End":"02:22.895","Text":"I can multiply both sides by x and get the final form,"},{"Start":"02:22.895 ","End":"02:27.630","Text":"which is what we\u0027re required to show and we are done."}],"ID":6278},{"Watched":false,"Name":"Exercise 17","Duration":"4m 7s","ChapterTopicVideoID":6265,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6265.jpeg","UploadDate":"2017-01-26T05:18:42.9230000","DurationForVideoObject":"PT4M7S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.650","Text":"Before I begin this exercise,"},{"Start":"00:01.650 ","End":"00:06.100","Text":"I\u0027d like to point out that there are actually 2 mean value theorems."},{"Start":"00:06.100 ","End":"00:11.640","Text":"The usual 1 that you\u0027re all familiar with is due to Lagrange,"},{"Start":"00:11.640 ","End":"00:17.790","Text":"but there is a more generalized 1 due to Cauchy generalized on Lagrange."},{"Start":"00:17.790 ","End":"00:20.850","Text":"If you haven\u0027t learned the Cauchy Mean Value Theorem"},{"Start":"00:20.850 ","End":"00:24.480","Text":"and you\u0027ll soon see what it is then you can skip this exercise."},{"Start":"00:24.480 ","End":"00:26.955","Text":"Having said that, let\u0027s get started."},{"Start":"00:26.955 ","End":"00:28.485","Text":"This is what I have to prove."},{"Start":"00:28.485 ","End":"00:30.360","Text":"I think there are 2 functions here."},{"Start":"00:30.360 ","End":"00:33.810","Text":"There\u0027s the arctangent and the natural logarithm and that immediately"},{"Start":"00:33.810 ","End":"00:37.760","Text":"hints that I\u0027m going to need the Cauchy Mean Value Theorem,"},{"Start":"00:37.760 ","End":"00:40.140","Text":"which relates to 2 separate functions."},{"Start":"00:40.140 ","End":"00:41.630","Text":"We have an f and a g."},{"Start":"00:41.630 ","End":"00:43.670","Text":"We let f be the arctangent,"},{"Start":"00:43.670 ","End":"00:46.820","Text":"g be the natural log of 1 plus x."},{"Start":"00:46.820 ","End":"00:48.470","Text":"They\u0027re both continuous"},{"Start":"00:48.470 ","End":"00:54.365","Text":"and differentiable on every x between 0 and 1, including the 0."},{"Start":"00:54.365 ","End":"00:56.660","Text":"Therefore, they\u0027re differentiable on any"},{"Start":"00:56.660 ","End":"01:01.030","Text":"interval 0, x, which is like a, b."},{"Start":"01:01.030 ","End":"01:04.170","Text":"According to the Cauchy Mean Value Theorem,"},{"Start":"01:04.170 ","End":"01:10.245","Text":"there is a c between 0 and x such that this equation holds."},{"Start":"01:10.245 ","End":"01:11.900","Text":"Now, this is where it\u0027s different from"},{"Start":"01:11.900 ","End":"01:13.760","Text":"the Lagrange Mean Value Theorem"},{"Start":"01:13.760 ","End":"01:16.535","Text":"because there we would have had x minus 0,"},{"Start":"01:16.535 ","End":"01:19.880","Text":"and here, we would have just had f prime of c."},{"Start":"01:19.880 ","End":"01:22.055","Text":"This doesn\u0027t look familiar?"},{"Start":"01:22.055 ","End":"01:23.900","Text":"You might want to skip this clip."},{"Start":"01:23.900 ","End":"01:26.070","Text":"Okay. But if you\u0027re staying with us,"},{"Start":"01:26.070 ","End":"01:27.560","Text":"then let me continue."},{"Start":"01:27.560 ","End":"01:30.205","Text":"We know what f and g are,"},{"Start":"01:30.205 ","End":"01:35.345","Text":"f is arctangent g is natural logarithm of 1 plus whatever it is."},{"Start":"01:35.345 ","End":"01:39.319","Text":"The derivatives are easily computable in our heads."},{"Start":"01:39.319 ","End":"01:42.140","Text":"Let\u0027s go ahead and compute all these."},{"Start":"01:42.140 ","End":"01:44.210","Text":"We\u0027ll get the following expression,"},{"Start":"01:44.210 ","End":"01:46.795","Text":"f is arctangent here and here,"},{"Start":"01:46.795 ","End":"01:50.775","Text":"g of x and g of 0, from here."},{"Start":"01:50.775 ","End":"01:55.970","Text":"The derivative of the arctangent of x is 1 over 1 plus x squared."},{"Start":"01:55.970 ","End":"01:57.830","Text":"But of course, here we substitute c."},{"Start":"01:57.830 ","End":"02:00.950","Text":"The derivative of the natural log of 1 plus x"},{"Start":"02:00.950 ","End":"02:04.520","Text":"is 1 over 1 plus x and we replace x by c"},{"Start":"02:04.520 ","End":"02:06.680","Text":"so we get this expression."},{"Start":"02:06.680 ","End":"02:09.455","Text":"Let\u0027s do a bit of simplification here."},{"Start":"02:09.455 ","End":"02:12.690","Text":"Arctangent of x, arctangent of 0 is"},{"Start":"02:12.690 ","End":"02:15.830","Text":"0 you either know this or you can do it on your calculator."},{"Start":"02:15.830 ","End":"02:17.480","Text":"It\u0027s shift tangent."},{"Start":"02:17.480 ","End":"02:19.219","Text":"Anyway, that\u0027s 0."},{"Start":"02:19.219 ","End":"02:21.875","Text":"Natural logarithm of 1 is 0."},{"Start":"02:21.875 ","End":"02:25.985","Text":"Here, when we have 1 over a divided by 1 over b."},{"Start":"02:25.985 ","End":"02:27.900","Text":"It\u0027s b over a."},{"Start":"02:27.900 ","End":"02:30.570","Text":"We end up with this expression."},{"Start":"02:30.570 ","End":"02:33.800","Text":"Here, we have a bit of a difficulty because we"},{"Start":"02:33.800 ","End":"02:37.060","Text":"would like to estimate what is this 1 plus c,"},{"Start":"02:37.060 ","End":"02:39.184","Text":"over 1 plus c squared."},{"Start":"02:39.184 ","End":"02:44.860","Text":"I mean, we know that c is between 0 and 1 because x is less than 1,"},{"Start":"02:44.860 ","End":"02:50.870","Text":"but the usual tricks don\u0027t work because c is an numerator, and denominator."},{"Start":"02:50.870 ","End":"02:53.150","Text":"There was a technique of trying to see whether"},{"Start":"02:53.150 ","End":"02:55.490","Text":"this is an increasing or decreasing function."},{"Start":"02:55.490 ","End":"03:03.380","Text":"Fortunately, here, there is a simple trick and that is that because c is between 0 and 1,"},{"Start":"03:03.380 ","End":"03:04.430","Text":"the reason for that is,"},{"Start":"03:04.430 ","End":"03:07.340","Text":"is because c is between 0 and x,"},{"Start":"03:07.340 ","End":"03:08.840","Text":"and x is less than 1."},{"Start":"03:08.840 ","End":"03:11.510","Text":"If c was, for example, a half,"},{"Start":"03:11.510 ","End":"03:15.635","Text":"then c squared is less than c because a quarter is less than a half,"},{"Start":"03:15.635 ","End":"03:19.255","Text":"a ninth is less than a third, and so on."},{"Start":"03:19.255 ","End":"03:25.010","Text":"What it means is that the numerator is bigger than the denominator."},{"Start":"03:25.010 ","End":"03:29.435","Text":"In this case, if the numerator is bigger than the denominator of positive numbers,"},{"Start":"03:29.435 ","End":"03:34.685","Text":"then we can say that this thing here is bigger than 1."},{"Start":"03:34.685 ","End":"03:37.415","Text":"That\u0027s all we need to continue."},{"Start":"03:37.415 ","End":"03:39.380","Text":"I don\u0027t even need an upper bound."},{"Start":"03:39.380 ","End":"03:43.745","Text":"It just need a lower bound for this because now I can say that"},{"Start":"03:43.745 ","End":"03:48.845","Text":"the arctangent of x over natural log of 1 plus x is bigger than 1."},{"Start":"03:48.845 ","End":"03:53.540","Text":"Then since the denominator is positive, I mean,"},{"Start":"03:53.540 ","End":"03:54.950","Text":"if x is bigger than 0,"},{"Start":"03:54.950 ","End":"03:56.360","Text":"1 plus x is bigger than 1,"},{"Start":"03:56.360 ","End":"03:58.865","Text":"and natural log of something bigger than 1 is positive."},{"Start":"03:58.865 ","End":"04:03.620","Text":"I multiply both sides by the denominator and get that this holds."},{"Start":"04:03.620 ","End":"04:06.335","Text":"This is exactly what we had to prove."},{"Start":"04:06.335 ","End":"04:08.640","Text":"So we are done."}],"ID":6279},{"Watched":false,"Name":"Exercise 18","Duration":"2m 3s","ChapterTopicVideoID":6266,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6266.jpeg","UploadDate":"2016-05-27T14:46:19.5700000","DurationForVideoObject":"PT2M3S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.050","Text":"Here we have another inequality to prove and once again,"},{"Start":"00:04.050 ","End":"00:08.175","Text":"we\u0027ll be using Lagrange\u0027s Mean Value Theorem."},{"Start":"00:08.175 ","End":"00:11.445","Text":"The function to take is the sine x"},{"Start":"00:11.445 ","End":"00:15.345","Text":"and it\u0027s continuous and differentiable just everywhere."},{"Start":"00:15.345 ","End":"00:18.750","Text":"Don\u0027t be confused by the notation here,"},{"Start":"00:18.750 ","End":"00:21.900","Text":"we are using x_2, x_1,"},{"Start":"00:21.900 ","End":"00:25.360","Text":"it might as well have been b and a,"},{"Start":"00:25.580 ","End":"00:29.350","Text":"the name shouldn\u0027t make a difference."},{"Start":"00:34.100 ","End":"00:38.625","Text":"What\u0027s going on here? There we are."},{"Start":"00:38.625 ","End":"00:47.895","Text":"So it satisfies the conditions of the Mean Value Theorem on x_1, x_2."},{"Start":"00:47.895 ","End":"00:53.720","Text":"So Lagrange\u0027s Mean Value Theorem guarantees the existence of a c"},{"Start":"00:53.720 ","End":"01:00.050","Text":"between the 2 of them and which satisfies the usual equation."},{"Start":"01:00.050 ","End":"01:07.050","Text":"But I\u0027ve already written it as sine instead of just f of x_1,"},{"Start":"01:07.050 ","End":"01:09.120","Text":"f of x_2, and so on."},{"Start":"01:09.120 ","End":"01:11.810","Text":"We know the derivative of sine is cosine so I cut"},{"Start":"01:11.810 ","End":"01:15.560","Text":"out a step here because you are sophisticated enough already."},{"Start":"01:17.060 ","End":"01:21.170","Text":"What we can do here is we can put this whole thing in"},{"Start":"01:21.170 ","End":"01:27.305","Text":"absolute value and the absolute value can go on top and bottom separately."},{"Start":"01:27.305 ","End":"01:34.155","Text":"Now, the cosine, as you know, oscillates between minus 1 and 1 never goes beyond."},{"Start":"01:34.155 ","End":"01:40.310","Text":"The absolute value is at most 1 and so what we can"},{"Start":"01:40.310 ","End":"01:42.920","Text":"get is that this thing is less than or equal to 1 because"},{"Start":"01:42.920 ","End":"01:47.300","Text":"the cosine can never exceed 1 in magnitude."},{"Start":"01:47.300 ","End":"01:51.290","Text":"All we have to do now to get to what we have to prove is just to"},{"Start":"01:51.290 ","End":"01:56.015","Text":"multiply both sides by this thing,"},{"Start":"01:56.015 ","End":"01:58.550","Text":"which is positive because x_1 is not equal to"},{"Start":"01:58.550 ","End":"02:03.480","Text":"x_2 and so we end up with this and we are done."}],"ID":6280},{"Watched":false,"Name":"Exercise 19","Duration":"2m ","ChapterTopicVideoID":6267,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6267.jpeg","UploadDate":"2017-01-26T05:19:20.0330000","DurationForVideoObject":"PT2M","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.270","Text":"Here we have to prove this inequality very similar to"},{"Start":"00:03.270 ","End":"00:06.960","Text":"one we did before with the sine and the cosine."},{"Start":"00:06.960 ","End":"00:08.730","Text":"Again, we\u0027re using x_2,"},{"Start":"00:08.730 ","End":"00:10.370","Text":"x_1 instead of b and a,"},{"Start":"00:10.370 ","End":"00:12.315","Text":"and that shouldn\u0027t matter at all."},{"Start":"00:12.315 ","End":"00:14.415","Text":"The key function here is the cosine."},{"Start":"00:14.415 ","End":"00:16.725","Text":"Let\u0027s see what\u0027s with the cosine."},{"Start":"00:16.725 ","End":"00:20.310","Text":"Well, the cosine is actually continuous and differentiable for all x,"},{"Start":"00:20.310 ","End":"00:23.385","Text":"it just keeps oscillating and so on to infinity."},{"Start":"00:23.385 ","End":"00:29.010","Text":"In particular, it will be so on the interval from x_1 to x_2."},{"Start":"00:29.010 ","End":"00:32.020","Text":"Let\u0027s assume that x_1 is not equal to x_2."},{"Start":"00:32.020 ","End":"00:34.550","Text":"It works if x_1 is equal to x_2,"},{"Start":"00:34.550 ","End":"00:36.145","Text":"because if x_1 is equal to x_2,"},{"Start":"00:36.145 ","End":"00:38.895","Text":"it just says that 0 is less than or equal to 0,"},{"Start":"00:38.895 ","End":"00:41.210","Text":"which is true but not very interesting."},{"Start":"00:41.210 ","End":"00:43.760","Text":"We\u0027ll assume that x_2 and x_1 are different."},{"Start":"00:43.760 ","End":"00:46.205","Text":"Anyway, according to the mean value theorem,"},{"Start":"00:46.205 ","End":"00:48.620","Text":"since we have continuity and differentiability,"},{"Start":"00:48.620 ","End":"00:51.985","Text":"we have this point c between x_1 and x_2."},{"Start":"00:51.985 ","End":"00:55.170","Text":"I guess we\u0027re even assuming here more than just that they\u0027re different."},{"Start":"00:55.170 ","End":"00:59.585","Text":"We\u0027ll assume that x_1 is actually less than x_2."},{"Start":"00:59.585 ","End":"01:02.255","Text":"I mean, 1 of them is got to be smaller if they\u0027re not equal."},{"Start":"01:02.255 ","End":"01:06.955","Text":"Let\u0027s label the smaller 1, x_1 and the larger 1, x_2."},{"Start":"01:06.955 ","End":"01:09.045","Text":"Now we\u0027re all set here."},{"Start":"01:09.045 ","End":"01:14.450","Text":"There exists such a c between the 2 of them which satisfies the usual equation."},{"Start":"01:14.450 ","End":"01:19.320","Text":"Only this time I haven\u0027t written it abstractly with f x_2 minus f of x_1 and so on."},{"Start":"01:19.320 ","End":"01:22.325","Text":"Have straight away written the function is cosine,"},{"Start":"01:22.325 ","End":"01:28.565","Text":"f prime of c is minus sine c because the derivative of cosine is minus sine."},{"Start":"01:28.565 ","End":"01:32.725","Text":"We proceed by taking the absolute value of both sides"},{"Start":"01:32.725 ","End":"01:37.115","Text":"because the absolute value of negative doesn\u0027t make much difference."},{"Start":"01:37.115 ","End":"01:41.840","Text":"We end up with absolute value of sine of c in the sine function,"},{"Start":"01:41.840 ","End":"01:43.400","Text":"like the cosine function,"},{"Start":"01:43.400 ","End":"01:45.350","Text":"is always less than or equal to 1."},{"Start":"01:45.350 ","End":"01:48.230","Text":"It oscillates between 1 and minus 1."},{"Start":"01:48.230 ","End":"01:49.825","Text":"Now that we have this,"},{"Start":"01:49.825 ","End":"01:52.250","Text":"and since x_2 is not equal to x_1,"},{"Start":"01:52.250 ","End":"01:53.660","Text":"well the absolute value is positive."},{"Start":"01:53.660 ","End":"01:57.485","Text":"I can multiply both sides by a positive and keep the inequality."},{"Start":"01:57.485 ","End":"02:01.920","Text":"We get this and that\u0027s what we had to prove. We are done."}],"ID":6281},{"Watched":false,"Name":"Exercise 20","Duration":"1m 49s","ChapterTopicVideoID":6268,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6268.jpeg","UploadDate":"2017-02-08T12:01:29.3930000","DurationForVideoObject":"PT1M49S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.050","Text":"In this exercise, we have to show that this inequality holds."},{"Start":"00:04.050 ","End":"00:07.260","Text":"Just notice that I forgot to write the conditions."},{"Start":"00:07.260 ","End":"00:10.950","Text":"The conditions are that x is bigger than 0,"},{"Start":"00:10.950 ","End":"00:13.410","Text":"but smaller than y."},{"Start":"00:13.410 ","End":"00:15.030","Text":"Let\u0027s get to the solution."},{"Start":"00:15.030 ","End":"00:18.524","Text":"We\u0027re going to, of course, use Lagrange\u0027s mean value theorem."},{"Start":"00:18.524 ","End":"00:21.885","Text":"The function we\u0027re going to take is the arc tangent function,"},{"Start":"00:21.885 ","End":"00:25.270","Text":"and it happens to be continuous and differentiable for all x,"},{"Start":"00:25.270 ","End":"00:27.330","Text":"from minus infinity to infinity."},{"Start":"00:27.330 ","End":"00:30.705","Text":"In particular, it\u0027s going to be continuous,"},{"Start":"00:30.705 ","End":"00:33.495","Text":"differentiable on the interval x, y,"},{"Start":"00:33.495 ","End":"00:36.435","Text":"so by the conclusions of Lagrange\u0027s theorem,"},{"Start":"00:36.435 ","End":"00:41.815","Text":"there is some c between x and y such that the following holds."},{"Start":"00:41.815 ","End":"00:43.940","Text":"Lagrange\u0027s theorem doesn\u0027t say it that way,"},{"Start":"00:43.940 ","End":"00:46.730","Text":"of course, we\u0027ve skipped a step, but really it says,"},{"Start":"00:46.730 ","End":"00:50.580","Text":"\"f of y and f of x,\" and here it says,"},{"Start":"00:50.580 ","End":"00:54.210","Text":"\"f prime of c,\" but we\u0027ve done so many of these,"},{"Start":"00:54.210 ","End":"00:56.130","Text":"I took some shortcuts here."},{"Start":"00:56.130 ","End":"00:58.385","Text":"As I say, f is arc tangent,"},{"Start":"00:58.385 ","End":"01:01.315","Text":"derivative of arc tangent and immediate integral."},{"Start":"01:01.315 ","End":"01:03.870","Text":"At the point c, it\u0027s 1 over 1 plus c squared."},{"Start":"01:03.870 ","End":"01:06.800","Text":"If I took it here, it would be 1 over 1 plus x squared."},{"Start":"01:06.800 ","End":"01:08.195","Text":"So that\u0027s settled."},{"Start":"01:08.195 ","End":"01:10.440","Text":"Next thing I want do is look,"},{"Start":"01:10.440 ","End":"01:11.700","Text":"there\u0027s absolute values here,"},{"Start":"01:11.700 ","End":"01:14.810","Text":"so the obvious thing to do is take absolute value here."},{"Start":"01:14.810 ","End":"01:16.685","Text":"This is what we get."},{"Start":"01:16.685 ","End":"01:18.960","Text":"Notice the right-hand side,"},{"Start":"01:18.960 ","End":"01:21.150","Text":"c is between x and y,"},{"Start":"01:21.150 ","End":"01:24.780","Text":"so certainly, c is bigger than 0,"},{"Start":"01:24.780 ","End":"01:27.695","Text":"and if c is bigger than 0,"},{"Start":"01:27.695 ","End":"01:31.085","Text":"then the denominator is strictly bigger than 1,"},{"Start":"01:31.085 ","End":"01:36.015","Text":"so 1 over something bigger than 1 is going to be less than 1."},{"Start":"01:36.015 ","End":"01:37.905","Text":"Now, y is bigger than x."},{"Start":"01:37.905 ","End":"01:39.030","Text":"This thing is positive."},{"Start":"01:39.030 ","End":"01:42.975","Text":"Multiply both sides by a positive quantity,"},{"Start":"01:42.975 ","End":"01:46.650","Text":"the inequality holds, and we get this."},{"Start":"01:46.650 ","End":"01:49.990","Text":"This is what we have to prove, so we\u0027re done."}],"ID":6282},{"Watched":false,"Name":"Exercise 21","Duration":"9m 23s","ChapterTopicVideoID":6269,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6269.jpeg","UploadDate":"2016-05-27T14:48:04.2200000","DurationForVideoObject":"PT9M23S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.210","Text":"Here we have another inequality to prove."},{"Start":"00:04.460 ","End":"00:08.340","Text":"It looks a bit strange because of the absolute values."},{"Start":"00:08.340 ","End":"00:13.870","Text":"Perhaps I should put a little dot in here to indicate that it\u0027s multiplied."},{"Start":"00:14.300 ","End":"00:20.235","Text":"First of all, an important remark because you may not want to listen to this clip."},{"Start":"00:20.235 ","End":"00:22.620","Text":"There are 2 mean value theorems."},{"Start":"00:22.620 ","End":"00:26.720","Text":"There\u0027s the regular 1, which is MVT,"},{"Start":"00:26.720 ","End":"00:32.420","Text":"mean value theorem, which is due to Lagrange, a French mathematician."},{"Start":"00:32.420 ","End":"00:37.355","Text":"There\u0027s a more generalized 1 due to Cauchy,"},{"Start":"00:37.355 ","End":"00:39.845","Text":"also a French mathematician."},{"Start":"00:39.845 ","End":"00:44.300","Text":"But he came later into the scene and he built on this 1 and generalize it."},{"Start":"00:44.300 ","End":"00:47.930","Text":"If you haven\u0027t learned the Cauchy mean value theorem"},{"Start":"00:47.930 ","End":"00:51.515","Text":"then you may not understand what\u0027s being said here."},{"Start":"00:51.515 ","End":"00:55.205","Text":"Well, if you\u0027re not sure just start and you\u0027ll see."},{"Start":"00:55.205 ","End":"00:59.000","Text":"This is usually characterized by having 2 different functions."},{"Start":"00:59.000 ","End":"01:01.910","Text":"Here, we have both a sine function and the tangent function."},{"Start":"01:01.910 ","End":"01:05.700","Text":"That\u0027s 1 of the giveaway is that it\u0027s going to be Cauchy."},{"Start":"01:08.090 ","End":"01:11.990","Text":"My second point is"},{"Start":"01:11.990 ","End":"01:16.880","Text":"that some of you may not be familiar with this notation for an interval."},{"Start":"01:16.880 ","End":"01:21.920","Text":"This symbol is the symbol belongs to."},{"Start":"01:21.920 ","End":"01:29.140","Text":"This says x, y belongs to the interval from 0 to Pi over 3 inclusive."},{"Start":"01:29.140 ","End":"01:36.220","Text":"If you don\u0027t like that I could say 0 less than or equal to,"},{"Start":"01:36.220 ","End":"01:40.140","Text":"often throw in 2 variables at once, x and y,"},{"Start":"01:40.140 ","End":"01:44.030","Text":"both of them less than or equal to Pi over 3,"},{"Start":"01:44.030 ","End":"01:47.430","Text":"so it makes you happier look at it that way."},{"Start":"01:47.530 ","End":"01:51.010","Text":"Now, I\u0027m going to properly begin."},{"Start":"01:51.010 ","End":"01:53.460","Text":"Like I said, there are 2 functions,"},{"Start":"01:53.460 ","End":"01:57.900","Text":"the tangent and the sine. Oh yeah."},{"Start":"01:57.900 ","End":"02:00.410","Text":"There was 1 other small thing that bothered me."},{"Start":"02:00.410 ","End":"02:01.970","Text":"It makes absolutely no difference,"},{"Start":"02:01.970 ","End":"02:05.085","Text":"but just to put things in good order,"},{"Start":"02:05.085 ","End":"02:07.320","Text":"here we had y before x."},{"Start":"02:07.320 ","End":"02:10.970","Text":"Here we should have really put the y here and the x here, of course,"},{"Start":"02:10.970 ","End":"02:14.090","Text":"it makes no difference because we have an absolute value and we can reverse"},{"Start":"02:14.090 ","End":"02:19.030","Text":"the terms but just say this is y and this is x."},{"Start":"02:19.100 ","End":"02:21.960","Text":"2 functions, tangent and sine,"},{"Start":"02:21.960 ","End":"02:30.570","Text":"they\u0027re both continuous and differentiable not on any interval but certainly."},{"Start":"02:30.570 ","End":"02:32.760","Text":"Then the interval x,"},{"Start":"02:32.760 ","End":"02:35.770","Text":"y between 0 and Pi over 3."},{"Start":"02:35.770 ","End":"02:37.700","Text":"As a matter of fact,"},{"Start":"02:37.700 ","End":"02:41.390","Text":"sine is continuous differentiable everywhere."},{"Start":"02:41.390 ","End":"02:50.280","Text":"Sorry, the tangent has the limitation of having an asymptote."},{"Start":"02:50.280 ","End":"02:54.635","Text":"It\u0027s not defined at Pi over 2 but we\u0027re still only up to Pi over 3."},{"Start":"02:54.635 ","End":"02:57.010","Text":"Long way to go till Pi over 2."},{"Start":"02:57.010 ","End":"03:01.415","Text":"This is 60 degrees and it\u0027s defined up to 90 if you like degrees."},{"Start":"03:01.415 ","End":"03:04.315","Text":"No problem with x, y."},{"Start":"03:04.315 ","End":"03:06.230","Text":"You\u0027re already used to having x,"},{"Start":"03:06.230 ","End":"03:08.720","Text":"y instead of a, b, and so on."},{"Start":"03:08.720 ","End":"03:16.565","Text":"The next thing is we\u0027ve satisfied the conditions of the mean value theorem."},{"Start":"03:16.565 ","End":"03:20.480","Text":"The conditions are essentially the same with Cauchy and Lagrange except"},{"Start":"03:20.480 ","End":"03:23.915","Text":"that with Cauchy there\u0027s 2 functions I have to satisfy it."},{"Start":"03:23.915 ","End":"03:29.030","Text":"Anyway, the conclusion is like before that there is"},{"Start":"03:29.030 ","End":"03:34.820","Text":"some value c between x and y but the formula is slightly different."},{"Start":"03:34.820 ","End":"03:37.840","Text":"Instead of just having y minus x on the bottom,"},{"Start":"03:37.840 ","End":"03:39.800","Text":"we have g of y minus g of x."},{"Start":"03:39.800 ","End":"03:43.310","Text":"Instead of having just f prime of c we have also a g prime of"},{"Start":"03:43.310 ","End":"03:47.385","Text":"c. Let\u0027s see what it means in our case."},{"Start":"03:47.385 ","End":"03:50.265","Text":"F is tangent, g is sine."},{"Start":"03:50.265 ","End":"03:55.995","Text":"What we get is tangent minus tangent, sine minus sine,"},{"Start":"03:55.995 ","End":"04:01.165","Text":"and the derivative of the tangent"},{"Start":"04:01.165 ","End":"04:06.679","Text":"is 1 over cosine squared and the argument is c. Similarly,"},{"Start":"04:06.679 ","End":"04:08.585","Text":"the derivative of sine is cosine,"},{"Start":"04:08.585 ","End":"04:11.585","Text":"so that goes on the denominator 2."},{"Start":"04:11.585 ","End":"04:15.270","Text":"Let me just give myself some more room here."},{"Start":"04:16.610 ","End":"04:19.520","Text":"We simplify the right-hand side."},{"Start":"04:19.520 ","End":"04:22.070","Text":"The cosine squared goes into the denominator and"},{"Start":"04:22.070 ","End":"04:26.150","Text":"combines with the cosine to form a cosine cubed."},{"Start":"04:26.150 ","End":"04:31.870","Text":"Now, we take the absolute value of both sides which we\u0027re allowed to do."},{"Start":"04:31.870 ","End":"04:38.600","Text":"Remember that c is between x and y,"},{"Start":"04:38.600 ","End":"04:41.675","Text":"but y is less than Pi over 3."},{"Start":"04:41.675 ","End":"04:47.850","Text":"What we can conclude from this is that c is less than, sorry."},{"Start":"04:50.270 ","End":"04:56.195","Text":"I was going to say there\u0027s a middle step here which really I should write it down."},{"Start":"04:56.195 ","End":"04:59.195","Text":"It follows, if I just look at part of this,"},{"Start":"04:59.195 ","End":"05:02.810","Text":"that c is less than Pi over 3,"},{"Start":"05:02.810 ","End":"05:04.790","Text":"means c is less than 60 degrees."},{"Start":"05:04.790 ","End":"05:12.590","Text":"Now, the cosine is a decreasing function from 0 to 90 degrees or from 0 to Pi over 2."},{"Start":"05:12.590 ","End":"05:14.634","Text":"If this is true,"},{"Start":"05:14.634 ","End":"05:20.340","Text":"that c is less than Pi over 3 and the cosine of c is bigger than the cosine of Pi over 3,"},{"Start":"05:20.340 ","End":"05:24.960","Text":"reverse the order of the inequality because cosine is decreasing."},{"Start":"05:25.790 ","End":"05:29.360","Text":"Now, if you look it up,"},{"Start":"05:29.360 ","End":"05:33.425","Text":"the cosine of 60 degrees is a half."},{"Start":"05:33.425 ","End":"05:36.230","Text":"Something that I have to memorize,"},{"Start":"05:36.230 ","End":"05:37.490","Text":"you can use your calculator,"},{"Start":"05:37.490 ","End":"05:39.350","Text":"look it up on tables."},{"Start":"05:39.350 ","End":"05:44.140","Text":"Cosine of 60 degrees or Pi over 3 is a half."},{"Start":"05:44.140 ","End":"05:50.360","Text":"Now, we want to get from here somehow to incorporate that into here."},{"Start":"05:50.360 ","End":"05:58.185","Text":"Cosine cubed raise both sides to the power of 3 with positive numbers, we\u0027re okay."},{"Start":"05:58.185 ","End":"06:01.450","Text":"Cosine cubed of c is bigger than an 1/8."},{"Start":"06:01.450 ","End":"06:04.595","Text":"Next thing is we can take the reciprocal."},{"Start":"06:04.595 ","End":"06:09.410","Text":"Reciprocal has the property of reversing the sign of the inequality."},{"Start":"06:09.410 ","End":"06:11.585","Text":"Reciprocal of 1/8 is 8."},{"Start":"06:11.585 ","End":"06:14.885","Text":"Reciprocal of the reciprocal is the thing itself."},{"Start":"06:14.885 ","End":"06:21.945","Text":"Now, if we take the absolute value of this."},{"Start":"06:21.945 ","End":"06:24.575","Text":"Well, we have the absolute value."},{"Start":"06:24.575 ","End":"06:27.770","Text":"Let me scroll back up. This was equal to this."},{"Start":"06:27.770 ","End":"06:34.875","Text":"We showed that this is less than 8."},{"Start":"06:34.875 ","End":"06:38.490","Text":"Actually, we skipped a step."},{"Start":"06:38.490 ","End":"06:43.280","Text":"If this is less than 8 then the absolute value of this, I mean it\u0027s positive."},{"Start":"06:43.280 ","End":"06:50.150","Text":"I could just put the absolute value here which doesn\u0027t"},{"Start":"06:50.150 ","End":"06:57.470","Text":"change anything but at least legitimizes that we\u0027ve replaced this by the tangent."},{"Start":"06:57.470 ","End":"06:59.480","Text":"Technical point."},{"Start":"06:59.480 ","End":"07:03.940","Text":"Finally, multiply both sides out by"},{"Start":"07:03.940 ","End":"07:09.510","Text":"the denominator which is positive because it\u0027s an absolute value."},{"Start":"07:09.820 ","End":"07:12.934","Text":"That is what we had to prove."},{"Start":"07:12.934 ","End":"07:20.220","Text":"Now, there is some delicacy which I omitted."},{"Start":"07:20.220 ","End":"07:27.170","Text":"You see, when I multiply by sine y minus sine x,"},{"Start":"07:27.170 ","End":"07:30.800","Text":"I\u0027m assuming that this thing is not 0."},{"Start":"07:30.800 ","End":"07:34.655","Text":"I mean, if it was 0 it wouldn\u0027t even be defined."},{"Start":"07:34.655 ","End":"07:44.660","Text":"In fact, yeah, this whole thing applies for proper intervals x and y which are,"},{"Start":"07:44.660 ","End":"07:46.645","Text":"let\u0027s go back up here,"},{"Start":"07:46.645 ","End":"07:49.415","Text":"x less than y."},{"Start":"07:49.415 ","End":"07:55.985","Text":"But the question didn\u0027t say anything about x being less than y."},{"Start":"07:55.985 ","End":"07:59.810","Text":"Let me just make this more refined."},{"Start":"07:59.810 ","End":"08:04.639","Text":"If x is less than y, we\u0027re okay."},{"Start":"08:04.639 ","End":"08:11.010","Text":"We can assume that x is the smaller of the 2 if they\u0027re not equal."},{"Start":"08:11.010 ","End":"08:14.135","Text":"Otherwise just call y x and call x y."},{"Start":"08:14.135 ","End":"08:17.960","Text":"But there is the other possibility that x equals y."},{"Start":"08:17.960 ","End":"08:21.770","Text":"This is the thing that we haven\u0027t really taken care of"},{"Start":"08:21.770 ","End":"08:27.765","Text":"properly and that is the possibility what if x equals y?"},{"Start":"08:27.765 ","End":"08:33.440","Text":"Then we don\u0027t have an interval x, y."},{"Start":"08:33.440 ","End":"08:36.590","Text":"We can\u0027t really continue and we got into trouble"},{"Start":"08:36.590 ","End":"08:40.805","Text":"back there we would be dividing by 0 and so on."},{"Start":"08:40.805 ","End":"08:44.150","Text":"But if x equals y,"},{"Start":"08:44.150 ","End":"08:45.890","Text":"tangent x equals tangent y,"},{"Start":"08:45.890 ","End":"08:47.870","Text":"and sine x equals sine y,"},{"Start":"08:47.870 ","End":"08:55.550","Text":"and essentially what we get is that 0 is less than or equal to 8 times 0 which is true."},{"Start":"08:55.550 ","End":"08:58.280","Text":"For x equals y it certainly holds true."},{"Start":"08:58.280 ","End":"08:59.810","Text":"Not only do we have less than or equal to,"},{"Start":"08:59.810 ","End":"09:01.115","Text":"we have a equal to."},{"Start":"09:01.115 ","End":"09:05.570","Text":"If x is not equal to y then we actually get that it"},{"Start":"09:05.570 ","End":"09:11.480","Text":"is not just less than or equal to, but less than."},{"Start":"09:11.480 ","End":"09:14.575","Text":"But either way, whether it\u0027s equal or whether it\u0027s less than,"},{"Start":"09:14.575 ","End":"09:16.375","Text":"it\u0027s less than or equal to."},{"Start":"09:16.375 ","End":"09:23.110","Text":"Anyway. I took care of that little technical point which was important and we\u0027re done."}],"ID":6283},{"Watched":false,"Name":"Exercise 22","Duration":"1m 43s","ChapterTopicVideoID":8309,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8309.jpeg","UploadDate":"2017-01-26T05:19:48.5900000","DurationForVideoObject":"PT1M43S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.010","Text":"In this exercise, we have to prove"},{"Start":"00:02.010 ","End":"00:05.420","Text":"this inequality and we\u0027ve got some actual numbers here,"},{"Start":"00:05.420 ","End":"00:06.749","Text":"they\u0027re not just letters."},{"Start":"00:06.749 ","End":"00:08.520","Text":"Why does Lagrange come into this?"},{"Start":"00:08.520 ","End":"00:10.920","Text":"It usually talks about the difference of something,"},{"Start":"00:10.920 ","End":"00:12.780","Text":"f of something minus f of something."},{"Start":"00:12.780 ","End":"00:14.280","Text":"Well, that\u0027s not difficult."},{"Start":"00:14.280 ","End":"00:16.440","Text":"All we have to do is rewrite the logarithm"},{"Start":"00:16.440 ","End":"00:18.750","Text":"of the quotient as the difference of the logs and"},{"Start":"00:18.750 ","End":"00:23.655","Text":"if I write this as natural log of 3 minus natural log of 2,"},{"Start":"00:23.655 ","End":"00:27.195","Text":"then it will look a lot more like Lagrange."},{"Start":"00:27.195 ","End":"00:30.930","Text":"We can let f of x be natural log of x."},{"Start":"00:30.930 ","End":"00:36.240","Text":"It\u0027s continuous and differentiable whenever x is positive and certainly, therefore,"},{"Start":"00:36.240 ","End":"00:38.410","Text":"on the interval from 2-3,"},{"Start":"00:38.410 ","End":"00:40.580","Text":"3 is our b and 2 is our a,"},{"Start":"00:40.580 ","End":"00:42.415","Text":"in this case here."},{"Start":"00:42.415 ","End":"00:46.220","Text":"Because it satisfies the conditions of the mean value theorem,"},{"Start":"00:46.220 ","End":"00:50.330","Text":"we can get the conclusion that there exists some c between 2 and 3,"},{"Start":"00:50.330 ","End":"00:54.365","Text":"which satisfies f of 3 minus f of 2 over 3 minus 2"},{"Start":"00:54.365 ","End":"00:58.860","Text":"is f prime at that point c. 3 minus 2 is 1."},{"Start":"00:58.860 ","End":"01:04.725","Text":"F prime of c is 1 over c because the derivative is 1 over x but here we have it at c,"},{"Start":"01:04.725 ","End":"01:06.525","Text":"so we get this."},{"Start":"01:06.525 ","End":"01:12.680","Text":"C is between 2 and 3 and so when we take the reciprocal, we get this."},{"Start":"01:12.680 ","End":"01:17.150","Text":"I took the reciprocal because I wanted 1 over c but then it changes the order."},{"Start":"01:17.150 ","End":"01:18.770","Text":"If something\u0027s between 2 and 3,"},{"Start":"01:18.770 ","End":"01:21.410","Text":"the reciprocals between a third and a half."},{"Start":"01:21.410 ","End":"01:24.830","Text":"Instead of 1 over c, I can put this expression here,"},{"Start":"01:24.830 ","End":"01:29.190","Text":"and that gives us this and 3 minus 2 is 1,"},{"Start":"01:29.190 ","End":"01:32.475","Text":"so this gives us this."},{"Start":"01:32.475 ","End":"01:37.190","Text":"The original form was to go back to the natural log of 3 over 2,"},{"Start":"01:37.190 ","End":"01:39.800","Text":"which we already said is equal to the difference of"},{"Start":"01:39.800 ","End":"01:44.130","Text":"the logs is the log of the quotient and we are done."}],"ID":8480},{"Watched":false,"Name":"Exercise 23","Duration":"2m 29s","ChapterTopicVideoID":8310,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8310.jpeg","UploadDate":"2017-01-26T05:20:35.6400000","DurationForVideoObject":"PT2M29S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.190","Text":"Here we have to prove this inequality."},{"Start":"00:02.190 ","End":"00:03.704","Text":"It\u0027s all with numbers."},{"Start":"00:03.704 ","End":"00:08.280","Text":"It doesn\u0027t look like the format for Lagrange\u0027s mean value theorem."},{"Start":"00:08.280 ","End":"00:13.290","Text":"But how about we subtract 1 from all the terms of this inequality?"},{"Start":"00:13.290 ","End":"00:15.990","Text":"If we did that, then we would get,"},{"Start":"00:15.990 ","End":"00:19.710","Text":"1 over twice the square root of 2,"},{"Start":"00:19.710 ","End":"00:21.390","Text":"and the 1 goes away."},{"Start":"00:21.390 ","End":"00:25.800","Text":"Less than the square root of 2 minus 1,"},{"Start":"00:25.800 ","End":"00:28.020","Text":"is less than 0.5,"},{"Start":"00:28.020 ","End":"00:30.255","Text":"I prefer to write it as 1.5."},{"Start":"00:30.255 ","End":"00:32.610","Text":"Now, all I have to do to get it to really look like"},{"Start":"00:32.610 ","End":"00:35.100","Text":"Lagrange is put a square root sign here."},{"Start":"00:35.100 ","End":"00:37.665","Text":"That\u0027s no problem. Square root of 1 is 1."},{"Start":"00:37.665 ","End":"00:44.975","Text":"That suggests that we take f of x to be the square root of x and a and b will be 1 and 2."},{"Start":"00:44.975 ","End":"00:46.490","Text":"That in fact is what we\u0027re going to do."},{"Start":"00:46.490 ","End":"00:49.070","Text":"The function f of x is square root of x is"},{"Start":"00:49.070 ","End":"00:52.880","Text":"continuous and differentiable and all positive x,"},{"Start":"00:52.880 ","End":"00:55.310","Text":"in particular between 1 and 2."},{"Start":"00:55.310 ","End":"00:57.455","Text":"By the mean value theorem,"},{"Start":"00:57.455 ","End":"01:01.670","Text":"Lagrange\u0027s version, there exists some number c between 1 and 2,"},{"Start":"01:01.670 ","End":"01:04.420","Text":"such that this is the usual equation,"},{"Start":"01:04.420 ","End":"01:07.815","Text":"but let\u0027s put in the square root instead of f and"},{"Start":"01:07.815 ","End":"01:09.410","Text":"remember that the derivative of"},{"Start":"01:09.410 ","End":"01:12.470","Text":"the square root of x is 1 over twice the square root of x."},{"Start":"01:12.470 ","End":"01:13.940","Text":"Only I didn\u0027t write x here,"},{"Start":"01:13.940 ","End":"01:15.620","Text":"I wrote c because that\u0027s what we want."},{"Start":"01:15.620 ","End":"01:18.500","Text":"Now, copying the inequality here,"},{"Start":"01:18.500 ","End":"01:20.075","Text":"I have to get an inequality."},{"Start":"01:20.075 ","End":"01:24.509","Text":"Instead of c, I need an inequality which involves"},{"Start":"01:24.509 ","End":"01:30.035","Text":"1 over twice the square root of c. What I\u0027m going to do is build up to it."},{"Start":"01:30.035 ","End":"01:32.165","Text":"First step, I\u0027ll take the square root."},{"Start":"01:32.165 ","End":"01:35.720","Text":"Now, square root preserves the order of magnitude positive numbers,"},{"Start":"01:35.720 ","End":"01:37.415","Text":"so no problem here."},{"Start":"01:37.415 ","End":"01:39.640","Text":"Next step is to multiply by 2."},{"Start":"01:39.640 ","End":"01:41.315","Text":"2 is a positive number,"},{"Start":"01:41.315 ","End":"01:44.405","Text":"again, preserves the direction of the inequality."},{"Start":"01:44.405 ","End":"01:47.200","Text":"But now we\u0027re going to take a reciprocal,"},{"Start":"01:47.200 ","End":"01:51.560","Text":"but the reciprocal has the property and we\u0027ve seen this before, that it reverses."},{"Start":"01:51.560 ","End":"01:54.545","Text":"The first shall be last and the last shall be first."},{"Start":"01:54.545 ","End":"01:58.265","Text":"Here\u0027s the 2 square root of 1 over here."},{"Start":"01:58.265 ","End":"02:02.260","Text":"The last is first, 2 square root of 2 is 2 square root of 2."},{"Start":"02:02.260 ","End":"02:05.765","Text":"Now let\u0027s do some numerical evaluation."},{"Start":"02:05.765 ","End":"02:08.420","Text":"This gives us 1.5."},{"Start":"02:08.420 ","End":"02:12.020","Text":"1 over twice square root of c is this but should have mentioned that this"},{"Start":"02:12.020 ","End":"02:16.505","Text":"is equal to square root of 2 minus 1,"},{"Start":"02:16.505 ","End":"02:19.070","Text":"because 2 minus 1 is 1."},{"Start":"02:19.070 ","End":"02:21.485","Text":"This is in fact what we had to prove."},{"Start":"02:21.485 ","End":"02:26.960","Text":"Not quite, but if I go back up and we proved the equivalent to what this was,"},{"Start":"02:26.960 ","End":"02:30.390","Text":"we proved this bit. Anyway, we\u0027re done."}],"ID":8481},{"Watched":false,"Name":"Exercise 25","Duration":"2m 53s","ChapterTopicVideoID":8311,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8311.jpeg","UploadDate":"2017-01-26T05:21:39.2570000","DurationForVideoObject":"PT2M53S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.750","Text":"Here we have another inequality to demonstrate and as usual,"},{"Start":"00:03.750 ","End":"00:06.615","Text":"we\u0027ll be using Lagrange\u0027s mean value theorem."},{"Start":"00:06.615 ","End":"00:09.210","Text":"Now we\u0027re expecting a difference here."},{"Start":"00:09.210 ","End":"00:13.455","Text":"Function will be arc sine but we need arc sine b minus arc sine a."},{"Start":"00:13.455 ","End":"00:15.915","Text":"Obviously we have to subtract the pi over 6."},{"Start":"00:15.915 ","End":"00:22.320","Text":"Now, pi over 6 is equivalent to 30 degrees and since we\u0027re talking about arc sine,"},{"Start":"00:22.320 ","End":"00:25.275","Text":"I want the angle whose arc sine is 30 degrees."},{"Start":"00:25.275 ","End":"00:26.340","Text":"If we took at it the other way,"},{"Start":"00:26.340 ","End":"00:29.610","Text":"the sine of 30 degrees is 1/2."},{"Start":"00:29.610 ","End":"00:33.375","Text":"That\u0027s a well-known sine which is 0.5."},{"Start":"00:33.375 ","End":"00:35.480","Text":"Because arc sine is the reverse,"},{"Start":"00:35.480 ","End":"00:39.770","Text":"we get the arc sine of 0.5 is pi over 6."},{"Start":"00:39.770 ","End":"00:44.329","Text":"In other words, what I\u0027m saying is that we can subtract pi over 6 everywhere."},{"Start":"00:44.329 ","End":"00:47.420","Text":"But whereas here I remove pi over 6,"},{"Start":"00:47.420 ","End":"00:49.160","Text":"and here I remove pi over 6."},{"Start":"00:49.160 ","End":"00:51.860","Text":"Here I removed the equivalent of pi over 6,"},{"Start":"00:51.860 ","End":"00:54.215","Text":"which is the arc sine of 0.5."},{"Start":"00:54.215 ","End":"00:57.040","Text":"This is equivalent to this and now we\u0027re going to prove this."},{"Start":"00:57.040 ","End":"00:59.585","Text":"In a previous exercise,"},{"Start":"00:59.585 ","End":"01:04.055","Text":"we showed that this inequality holds in general for b and"},{"Start":"01:04.055 ","End":"01:09.380","Text":"a and because we have this general result where a and b are between 0 and 1,"},{"Start":"01:09.380 ","End":"01:10.610","Text":"which is the case here."},{"Start":"01:10.610 ","End":"01:15.905","Text":"I mean, certainly 0.5 and 0.6 are between 0 and 1."},{"Start":"01:15.905 ","End":"01:19.745","Text":"It\u0027s okay if I take b to be this a to be that,"},{"Start":"01:19.745 ","End":"01:21.215","Text":"and if I substitute,"},{"Start":"01:21.215 ","End":"01:27.770","Text":"then this thing just translates as this just replacing everything b 0.68, a 0.5."},{"Start":"01:27.770 ","End":"01:29.255","Text":"Straightforward substitution."},{"Start":"01:29.255 ","End":"01:31.435","Text":"Now a little bit of computation."},{"Start":"01:31.435 ","End":"01:33.215","Text":"I\u0027ll write down what we get."},{"Start":"01:33.215 ","End":"01:35.825","Text":"This minus this is 0.1,"},{"Start":"01:35.825 ","End":"01:37.430","Text":"1/2 squared is 1/4."},{"Start":"01:37.430 ","End":"01:41.670","Text":"1 minus 1/4 is 3/4 again, 0.1,"},{"Start":"01:41.670 ","End":"01:44.580","Text":"0.6 squared is 0.36,"},{"Start":"01:44.580 ","End":"01:49.830","Text":"1 minus 0.36 is 0.64 square root of 0.64, 0.8."},{"Start":"01:49.830 ","End":"01:52.415","Text":"Okay. We\u0027re up to here now,"},{"Start":"01:52.415 ","End":"01:54.150","Text":"a little bit more canceling."},{"Start":"01:54.150 ","End":"01:56.630","Text":"This is the square root of 3 over the square root of"},{"Start":"01:56.630 ","End":"02:00.470","Text":"4 and 1 over the square root of 4 is 1/2,"},{"Start":"02:00.470 ","End":"02:04.460","Text":"and when I take the 2 to the numerator, it becomes 0.2."},{"Start":"02:04.460 ","End":"02:08.750","Text":"That\u0027s that. The other thing is that if I multiply top and bottom by 10 here,"},{"Start":"02:08.750 ","End":"02:11.630","Text":"0.1 over 0.8 is the same as 1 over 8,"},{"Start":"02:11.630 ","End":"02:15.770","Text":"and we had a square root of 3 in the numerator."},{"Start":"02:15.770 ","End":"02:21.645","Text":"The usual trick is to multiply by square root of 3 over the square root of 3 and"},{"Start":"02:21.645 ","End":"02:24.860","Text":"the other thing I\u0027m going to do is try and get rid of this decimal by"},{"Start":"02:24.860 ","End":"02:28.460","Text":"multiplying it by 5 and then 5 in the bottom also."},{"Start":"02:28.460 ","End":"02:37.085","Text":"Let\u0027s see, 5 with 0.2 gives me 1 and square root of 3 with square root of 3 together,"},{"Start":"02:37.085 ","End":"02:41.840","Text":"give me 3 and 3 times 5 is 15."},{"Start":"02:41.840 ","End":"02:46.040","Text":"Altogether, what we get is that here we have the square root of 3."},{"Start":"02:46.040 ","End":"02:48.095","Text":"The 3 times 5 is 15,"},{"Start":"02:48.095 ","End":"02:51.590","Text":"and the rest of it is the same and this is what we had to prove."},{"Start":"02:51.590 ","End":"02:53.850","Text":"We are done."}],"ID":8482},{"Watched":false,"Name":"Exercise 24","Duration":"4m 9s","ChapterTopicVideoID":6272,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6272.jpeg","UploadDate":"2017-02-08T12:02:53.9170000","DurationForVideoObject":"PT4M9S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.275","Text":"This exercise, we have to prove the following inequality, so with numbers,"},{"Start":"00:04.275 ","End":"00:06.540","Text":"not too many letters, Pi,"},{"Start":"00:06.540 ","End":"00:09.420","Text":"and we want to use Lagrange\u0027s mean value theorem."},{"Start":"00:09.420 ","End":"00:12.885","Text":"But for Lagrange, we expect to have a difference here."},{"Start":"00:12.885 ","End":"00:16.814","Text":"If I subtract Pi over 4 from all of these,"},{"Start":"00:16.814 ","End":"00:20.160","Text":"the reason I\u0027m going to subtract Pi over 4, well, for several reasons."},{"Start":"00:20.160 ","End":"00:24.030","Text":"First of all, I have a difference and also the Pi over 4 happens"},{"Start":"00:24.030 ","End":"00:28.275","Text":"to be the arctangent of 1."},{"Start":"00:28.275 ","End":"00:30.390","Text":"You can do this on your calculator,"},{"Start":"00:30.390 ","End":"00:32.180","Text":"shift tangent of 1."},{"Start":"00:32.180 ","End":"00:35.690","Text":"But I happen to know that tangent of 45 degrees,"},{"Start":"00:35.690 ","End":"00:40.549","Text":"this is 1 of those basic angles is 1 and 45 degrees is Pi over 4."},{"Start":"00:40.549 ","End":"00:44.285","Text":"Refresh your memory, we have a 45 degree right triangle."},{"Start":"00:44.285 ","End":"00:47.270","Text":"The tangent of 45 degrees is this over this,"},{"Start":"00:47.270 ","End":"00:49.205","Text":"so this equals this, so it\u0027s 1 and so on."},{"Start":"00:49.205 ","End":"00:51.700","Text":"This is the thing I\u0027m going to use."},{"Start":"00:51.700 ","End":"00:54.424","Text":"We\u0027re going to rewrite this in an equivalent form,"},{"Start":"00:54.424 ","End":"00:58.175","Text":"taking away Pi over 4 from here and from here,"},{"Start":"00:58.175 ","End":"01:01.264","Text":"and from here, we\u0027ll look at the inequality."},{"Start":"01:01.264 ","End":"01:04.115","Text":"Only when I took the Pi over 4 from here,"},{"Start":"01:04.115 ","End":"01:06.440","Text":"I wrote it as arctangent of 1."},{"Start":"01:06.440 ","End":"01:08.420","Text":"Now it looks very much like"},{"Start":"01:08.420 ","End":"01:14.270","Text":"Lagrange\u0027s theorem setup because if I let f of x equals arctangent x,"},{"Start":"01:14.270 ","End":"01:20.055","Text":"and I look at the interval from 1 to 4/3, then it\u0027s differentiable."},{"Start":"01:20.055 ","End":"01:22.070","Text":"By the mean value theorem,"},{"Start":"01:22.070 ","End":"01:28.100","Text":"we have some c that is between 1 and 4/3 and satisfies the usual equation."},{"Start":"01:28.100 ","End":"01:30.305","Text":"But do it without the f part,"},{"Start":"01:30.305 ","End":"01:33.260","Text":"f of 4/3 minus f of 1,"},{"Start":"01:33.260 ","End":"01:37.225","Text":"which is the arctangent over 4/3 minus 1."},{"Start":"01:37.225 ","End":"01:39.680","Text":"This is like my a and b."},{"Start":"01:39.680 ","End":"01:42.695","Text":"This is a and this is b, the 4/3."},{"Start":"01:42.695 ","End":"01:44.480","Text":"Anything that needs explaining is this,"},{"Start":"01:44.480 ","End":"01:47.570","Text":"but not really because it\u0027s an immediate derivative."},{"Start":"01:47.570 ","End":"01:51.815","Text":"The derivative of arctangent x is 1/1 plus x squared,"},{"Start":"01:51.815 ","End":"01:53.090","Text":"but we don\u0027t want it for x,"},{"Start":"01:53.090 ","End":"01:59.330","Text":"we want it for c. This is the f prime of c. This is the f of 4/3,"},{"Start":"01:59.330 ","End":"02:02.640","Text":"f of a minus f of b over b minus a."},{"Start":"02:02.640 ","End":"02:05.615","Text":"Now we need to talk about inequalities."},{"Start":"02:05.615 ","End":"02:07.680","Text":"The inequality that we have,"},{"Start":"02:07.680 ","End":"02:10.215","Text":"meanwhile, is this inequality."},{"Start":"02:10.215 ","End":"02:12.315","Text":"I\u0027m going to just copy over here."},{"Start":"02:12.315 ","End":"02:17.555","Text":"I want to somehow get from c to 1/1 plus c squared."},{"Start":"02:17.555 ","End":"02:19.040","Text":"Let\u0027s do that in steps."},{"Start":"02:19.040 ","End":"02:23.900","Text":"The first thing we can do is raise everything to the power of 2 or square everything."},{"Start":"02:23.900 ","End":"02:28.355","Text":"Squaring positive numbers doesn\u0027t change the order of the inequality."},{"Start":"02:28.355 ","End":"02:31.715","Text":"The next thing we can do is add 1 to everything."},{"Start":"02:31.715 ","End":"02:33.010","Text":"We get 1 plus this,"},{"Start":"02:33.010 ","End":"02:35.000","Text":"1 plus this, 1 plus this."},{"Start":"02:35.000 ","End":"02:37.880","Text":"Then we take the reciprocal."},{"Start":"02:37.880 ","End":"02:39.935","Text":"But when we take the reciprocal,"},{"Start":"02:39.935 ","End":"02:41.600","Text":"we invert the order."},{"Start":"02:41.600 ","End":"02:44.165","Text":"The middle term is 1/1 plus c squared."},{"Start":"02:44.165 ","End":"02:47.945","Text":"The first term is 1 over all this."},{"Start":"02:47.945 ","End":"02:50.375","Text":"This is 1 plus 16/9,"},{"Start":"02:50.375 ","End":"02:56.225","Text":"which gives us 25/9 and 1 plus 1,"},{"Start":"02:56.225 ","End":"02:59.450","Text":"we don\u0027t need a calculator for that is 2."},{"Start":"02:59.450 ","End":"03:06.670","Text":"When I do the reciprocal that 25/9 becomes 9/25 and the 2 becomes 1/2."},{"Start":"03:06.670 ","End":"03:09.905","Text":"Now, we\u0027ve got this and all I have to do is replace,"},{"Start":"03:09.905 ","End":"03:12.080","Text":"you see the 1/1 plus c squared, c,"},{"Start":"03:12.080 ","End":"03:13.475","Text":"1/1 plus c squared,"},{"Start":"03:13.475 ","End":"03:15.845","Text":"just replace this in here."},{"Start":"03:15.845 ","End":"03:19.835","Text":"What we get is that 9/25 and the 1/2,"},{"Start":"03:19.835 ","End":"03:22.880","Text":"and I\u0027ve replaced the 1/1 plus c squared,"},{"Start":"03:22.880 ","End":"03:24.950","Text":"like I said, from here."},{"Start":"03:24.950 ","End":"03:27.965","Text":"The rest of it is just simple arithmetic."},{"Start":"03:27.965 ","End":"03:30.605","Text":"4/3 minus 1 is 1/3."},{"Start":"03:30.605 ","End":"03:34.505","Text":"The 1/3 I\u0027m going to multiply both here and here."},{"Start":"03:34.505 ","End":"03:38.345","Text":"That gives us what we originally wanted to prove almost."},{"Start":"03:38.345 ","End":"03:41.360","Text":"We just have to do a bit of cancellation."},{"Start":"03:41.360 ","End":"03:44.650","Text":"This 3 goes into this 9,"},{"Start":"03:44.650 ","End":"03:51.150","Text":"3 times and 1/2 times 1/3 is 1/6."},{"Start":"03:51.150 ","End":"03:53.130","Text":"Here we have 3/25,"},{"Start":"03:53.130 ","End":"03:54.685","Text":"and here we have 1/6."},{"Start":"03:54.685 ","End":"03:57.920","Text":"This is exactly what is written here,"},{"Start":"03:57.920 ","End":"04:01.580","Text":"which is equivalent to we said to what the original thing was,"},{"Start":"04:01.580 ","End":"04:04.625","Text":"3/25, 1/6, check again,"},{"Start":"04:04.625 ","End":"04:06.815","Text":"3/25 and here 1/6."},{"Start":"04:06.815 ","End":"04:09.510","Text":"That is right, we are done."}],"ID":6286},{"Watched":false,"Name":"Exercise 26","Duration":"3m 3s","ChapterTopicVideoID":8312,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8312.jpeg","UploadDate":"2017-01-26T05:22:39.6130000","DurationForVideoObject":"PT3M3S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.850","Text":"This exercise is a little bit more theoretical."},{"Start":"00:02.850 ","End":"00:05.760","Text":"We\u0027re given a function which is differentiable for all x,"},{"Start":"00:05.760 ","End":"00:08.310","Text":"and I want to point out that if it\u0027s differentiable for all x,"},{"Start":"00:08.310 ","End":"00:12.390","Text":"it\u0027s also continuous for all x but you can just assume that we were told that it\u0027s"},{"Start":"00:12.390 ","End":"00:16.650","Text":"continuous for all x and it satisfies this inequality for its derivative."},{"Start":"00:16.650 ","End":"00:20.565","Text":"Its derivative is bounded by 5 in absolute value."},{"Start":"00:20.565 ","End":"00:25.680","Text":"We also have some further information that we know what f of 1 is and"},{"Start":"00:25.680 ","End":"00:31.920","Text":"what f of 4 is and we have to show that f of 2 is 8."},{"Start":"00:31.920 ","End":"00:34.710","Text":"Let me remind you of what Lagrange\u0027s theorem says,"},{"Start":"00:34.710 ","End":"00:36.435","Text":"that\u0027s the mean value theorem."},{"Start":"00:36.435 ","End":"00:39.090","Text":"It says that under certain conditions,"},{"Start":"00:39.090 ","End":"00:40.500","Text":"that this equation holds."},{"Start":"00:40.500 ","End":"00:47.100","Text":"F of b minus f of a over b minus a is equal to f prime of c for some c between a and b."},{"Start":"00:47.100 ","End":"00:52.665","Text":"This applies when f is continuous and differentiable on the interval a b."},{"Start":"00:52.665 ","End":"00:54.150","Text":"Now in our case,"},{"Start":"00:54.150 ","End":"00:56.925","Text":"we have an f, but what are a and b?"},{"Start":"00:56.925 ","End":"00:58.275","Text":"Well, the thing is,"},{"Start":"00:58.275 ","End":"01:01.440","Text":"we\u0027re given f of 1, we\u0027re given f of 4,"},{"Start":"01:01.440 ","End":"01:04.155","Text":"and we have to demonstrate something about f of 2,"},{"Start":"01:04.155 ","End":"01:08.745","Text":"and actually what we\u0027re going to do is apply Lagrange\u0027s theorem twice."},{"Start":"01:08.745 ","End":"01:10.500","Text":"See 2 is between 1 and 4,"},{"Start":"01:10.500 ","End":"01:15.260","Text":"we\u0027re going to apply it separately on the interval from 1-2,"},{"Start":"01:15.260 ","End":"01:19.565","Text":"then we\u0027re going to apply it on the interval from 2-4"},{"Start":"01:19.565 ","End":"01:24.080","Text":"altogether covering the interval from 1-4 but we\u0027re going to do it in 2 stages."},{"Start":"01:24.080 ","End":"01:27.960","Text":"Let\u0027s go to stage 1 and just for a simplicity,"},{"Start":"01:27.960 ","End":"01:33.125","Text":"let\u0027s say that f of 2 is k. What we get from Lagrange\u0027s theorem,"},{"Start":"01:33.125 ","End":"01:38.570","Text":"plugging in 4 and 2 here we get this where C1 is between 2 and 4,"},{"Start":"01:38.570 ","End":"01:42.320","Text":"I called it C1 and not just C because that\u0027s part 1 and in part 2,"},{"Start":"01:42.320 ","End":"01:43.715","Text":"we\u0027re going to call it C2."},{"Start":"01:43.715 ","End":"01:49.775","Text":"What this tells us is that since f of 2 is k and f of 4 is 18,"},{"Start":"01:49.775 ","End":"01:53.580","Text":"18 minus k over 2 is f prime of C,"},{"Start":"01:53.580 ","End":"01:57.545","Text":"and f prime of C is going to be less than or equal to"},{"Start":"01:57.545 ","End":"02:04.310","Text":"5 so we get that 18 minus k over 2 is less than or equal to 5."},{"Start":"02:04.310 ","End":"02:06.680","Text":"If I multiply both sides by 2,"},{"Start":"02:06.680 ","End":"02:13.045","Text":"which is positive 18 minus k less than 10 we get that k is bigger or equal to 8."},{"Start":"02:13.045 ","End":"02:14.465","Text":"That\u0027s part 1."},{"Start":"02:14.465 ","End":"02:17.150","Text":"Like I said, we\u0027re going to do 2 parts and what we\u0027re going to"},{"Start":"02:17.150 ","End":"02:20.660","Text":"get is that f of 2 minus f of"},{"Start":"02:20.660 ","End":"02:27.305","Text":"1 over 2 minus 1 is f prime at some other C and this time the C2 is between 1 and 2."},{"Start":"02:27.305 ","End":"02:31.730","Text":"But we know that f of 2 is k and f of 1 was given to be 3,"},{"Start":"02:31.730 ","End":"02:34.115","Text":"so we get this equality,"},{"Start":"02:34.115 ","End":"02:39.470","Text":"but f prime of every C is less than or equal to 5."},{"Start":"02:39.470 ","End":"02:41.870","Text":"We get that k minus 3 is less than or equal to 5,"},{"Start":"02:41.870 ","End":"02:45.005","Text":"adding 3, we get k is less than or equal to 8."},{"Start":"02:45.005 ","End":"02:51.995","Text":"Now look, we have that k is bigger or equal to 8 and k is less than or equal to 8."},{"Start":"02:51.995 ","End":"02:54.950","Text":"Well, if the only way it can be bigger or equal to and less or equal"},{"Start":"02:54.950 ","End":"02:58.655","Text":"to is if it\u0027s actually equal to and k,"},{"Start":"02:58.655 ","End":"03:02.090","Text":"after all, is f of 2 so this is what we had to show,"},{"Start":"03:02.090 ","End":"03:04.500","Text":"and we are done."}],"ID":8483},{"Watched":false,"Name":"Exercise 27","Duration":"2m 48s","ChapterTopicVideoID":8313,"CourseChapterTopicPlaylistID":84681,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8313.jpeg","UploadDate":"2017-01-26T05:23:33.3870000","DurationForVideoObject":"PT2M48S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.210","Text":"This exercise is similar to the previous 1."},{"Start":"00:03.210 ","End":"00:06.300","Text":"We\u0027re given a function differentiable for all x,"},{"Start":"00:06.300 ","End":"00:08.880","Text":"which means that it\u0027s also continuous for all x."},{"Start":"00:08.880 ","End":"00:12.120","Text":"Its derivative satisfies this inequality."},{"Start":"00:12.120 ","End":"00:15.510","Text":"We\u0027re also given f of 1 and f of 4 and we have to"},{"Start":"00:15.510 ","End":"00:19.455","Text":"show that f of 2 satisfies this inequality."},{"Start":"00:19.455 ","End":"00:21.915","Text":"We\u0027re going to use Lagrange\u0027s mean value theorem,"},{"Start":"00:21.915 ","End":"00:24.555","Text":"although it\u0027s not immediately apparent how."},{"Start":"00:24.555 ","End":"00:28.565","Text":"Let me write the Lagrange mean value theorem here."},{"Start":"00:28.565 ","End":"00:31.610","Text":"This is the conclusion that if f is continuous"},{"Start":"00:31.610 ","End":"00:35.330","Text":"on the interval from a to b and differentiable and so on and so on,"},{"Start":"00:35.330 ","End":"00:39.200","Text":"then we get that there is some c between a and b and this holds."},{"Start":"00:39.200 ","End":"00:41.720","Text":"As before, we\u0027re going to do it in 2 steps."},{"Start":"00:41.720 ","End":"00:43.429","Text":"The a and b are going to vary."},{"Start":"00:43.429 ","End":"00:47.030","Text":"We\u0027re going to break up the interval c. We were given the value of 1,"},{"Start":"00:47.030 ","End":"00:51.340","Text":"the value at 4, and we had to make some estimate on the value at 2."},{"Start":"00:51.340 ","End":"00:55.940","Text":"We\u0027re going to break up the interval from 1-4 separately as"},{"Start":"00:55.940 ","End":"01:01.590","Text":"the interval from 1-2 and from 2-4."},{"Start":"01:01.590 ","End":"01:03.830","Text":"I\u0027m going to apply Lagrange\u0027s theorem twice,"},{"Start":"01:03.830 ","End":"01:06.005","Text":"once with a is 1, b is 2,"},{"Start":"01:06.005 ","End":"01:08.690","Text":"and once with a is 2, b is 4."},{"Start":"01:08.690 ","End":"01:10.595","Text":"Let\u0027s do the first part."},{"Start":"01:10.595 ","End":"01:12.320","Text":"To make things simple,"},{"Start":"01:12.320 ","End":"01:14.140","Text":"because we\u0027re going to use f of 2 a lot,"},{"Start":"01:14.140 ","End":"01:22.610","Text":"we\u0027ll just call it k. This will be the first and this will be the second."},{"Start":"01:22.610 ","End":"01:23.855","Text":"We\u0027ll start with the 4,2."},{"Start":"01:23.855 ","End":"01:28.460","Text":"So f of 4 minus f of 2 over 4 minus 2 is f prime of some c,"},{"Start":"01:28.460 ","End":"01:29.820","Text":"I\u0027ll call it c_1,"},{"Start":"01:29.820 ","End":"01:31.080","Text":"which is between 2 and 4."},{"Start":"01:31.080 ","End":"01:33.270","Text":"F of 4 is given to be 18,"},{"Start":"01:33.270 ","End":"01:36.900","Text":"f of 2 is k, 4 minus 2 is 2,"},{"Start":"01:36.900 ","End":"01:40.140","Text":"and so this is what we get f prime of"},{"Start":"01:40.140 ","End":"01:44.025","Text":"all c is less than or equal to 7, this is what I get."},{"Start":"01:44.025 ","End":"01:46.100","Text":"Now to applying by 2,"},{"Start":"01:46.100 ","End":"01:48.410","Text":"got ahead of myself there and that\u0027s where we get"},{"Start":"01:48.410 ","End":"01:51.815","Text":"that 18 minus k less than or equal to 7 times 2 is 14,"},{"Start":"01:51.815 ","End":"01:55.220","Text":"which means that k is bigger or equal to 4."},{"Start":"01:55.220 ","End":"01:58.085","Text":"Bring the k here, 14 from 18 is 4,"},{"Start":"01:58.085 ","End":"02:03.985","Text":"and that\u0027s the first half of what we had to prove and I\u0027d like to highlight this."},{"Start":"02:03.985 ","End":"02:09.005","Text":"Now, continuing to the second bit with the interval from 1-2,"},{"Start":"02:09.005 ","End":"02:15.775","Text":"what we get is that f of 2 minus f of 1 over 2 minus 1 is f prime of a different c,"},{"Start":"02:15.775 ","End":"02:17.535","Text":"this time between 1-2,"},{"Start":"02:17.535 ","End":"02:23.745","Text":"but we were given f of 1 to be 3 and f of 2 we called it k and 2 minus 1 is 1."},{"Start":"02:23.745 ","End":"02:28.360","Text":"What we get is that k minus 3 is less than or equal to 7,"},{"Start":"02:28.360 ","End":"02:31.190","Text":"because f prime of all c is less than or equal to 7."},{"Start":"02:31.190 ","End":"02:34.070","Text":"Adding 3, we get that k is less than or equal to 10,"},{"Start":"02:34.070 ","End":"02:36.770","Text":"which I would also like to highlight."},{"Start":"02:36.770 ","End":"02:39.770","Text":"Now look, if we have k bigger or equal to 4,"},{"Start":"02:39.770 ","End":"02:40.940","Text":"and k less than or equal to 10,"},{"Start":"02:40.940 ","End":"02:44.240","Text":"I can just combine these 2 inequalities into 1 inequality"},{"Start":"02:44.240 ","End":"02:48.690","Text":"and write it this way and this is what we have to prove. So we\u0027re done."}],"ID":8484}],"Thumbnail":null,"ID":84681},{"Name":"Rolle`s Theorem","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Rolle`s Theorem","Duration":"14m 2s","ChapterTopicVideoID":6281,"CourseChapterTopicPlaylistID":84682,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6281.jpeg","UploadDate":"2019-11-14T07:19:07.0630000","DurationForVideoObject":"PT14M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.420","Text":"In this clip, we\u0027ll learn about Rolle\u0027s theorem,"},{"Start":"00:03.420 ","End":"00:09.125","Text":"1 of the great theorems in calculus and dates back to 1691."},{"Start":"00:09.125 ","End":"00:11.950","Text":"We\u0027re given a function."},{"Start":"00:11.950 ","End":"00:14.235","Text":"I\u0027ve written something down."},{"Start":"00:14.235 ","End":"00:17.490","Text":"We\u0027re given a function f, which satisfies certain conditions,"},{"Start":"00:17.490 ","End":"00:19.155","Text":"and I\u0027ll write them in a moment."},{"Start":"00:19.155 ","End":"00:21.630","Text":"If it satisfies these conditions,"},{"Start":"00:21.630 ","End":"00:25.710","Text":"then we get a certain conclusion."},{"Start":"00:25.710 ","End":"00:29.070","Text":"What are the conditions that the function satisfies?"},{"Start":"00:29.070 ","End":"00:35.980","Text":"First of all, that it\u0027s continuous on a closed interval, a,b."},{"Start":"00:36.080 ","End":"00:39.450","Text":"I hope you know what a closed interval is."},{"Start":"00:39.450 ","End":"00:44.225","Text":"It just means that for x between a and b,"},{"Start":"00:44.225 ","End":"00:47.850","Text":"including the end points a and b."},{"Start":"00:49.520 ","End":"00:57.710","Text":"Condition 2, f is differentiable on a,"},{"Start":"00:57.710 ","End":"00:59.795","Text":"b, the open interval,"},{"Start":"00:59.795 ","End":"01:04.370","Text":"meaning it may or may not be differentiable at the end points."},{"Start":"01:04.370 ","End":"01:06.515","Text":"I don\u0027t require it in any event."},{"Start":"01:06.515 ","End":"01:10.555","Text":"This of course means that x is between a and b, not inclusive."},{"Start":"01:10.555 ","End":"01:14.315","Text":"The third condition is that it\u0027s got to be equal at the endpoints."},{"Start":"01:14.315 ","End":"01:16.430","Text":"Other words, f of a,"},{"Start":"01:16.430 ","End":"01:17.960","Text":"is equal to f of b,"},{"Start":"01:17.960 ","End":"01:20.350","Text":"the same value at both ends."},{"Start":"01:20.350 ","End":"01:25.425","Text":"Now the conclusion from all these 3 conditions is that,"},{"Start":"01:25.425 ","End":"01:30.890","Text":"informally there\u0027s a point between a and b where the derivative is 0."},{"Start":"01:30.890 ","End":"01:34.890","Text":"There exists a point c,"},{"Start":"01:34.890 ","End":"01:37.845","Text":"such that c is between a and b,"},{"Start":"01:37.845 ","End":"01:40.330","Text":"such that,"},{"Start":"01:42.260 ","End":"01:50.290","Text":"f prime of c is equal to 0."},{"Start":"01:50.290 ","End":"01:54.140","Text":"I would like to repeat that, and then I\u0027ll draw a sketch and explain it."},{"Start":"01:54.140 ","End":"01:57.290","Text":"We have a function which satisfies 3 conditions."},{"Start":"01:57.290 ","End":"02:01.670","Text":"It\u0027s continuous on the closed interval a, b,"},{"Start":"02:01.670 ","End":"02:04.550","Text":"and it\u0027s differentiable on the open interval a,"},{"Start":"02:04.550 ","End":"02:10.355","Text":"b and the values of the end points are equal."},{"Start":"02:10.355 ","End":"02:15.395","Text":"In this case, there is some point between a and b,"},{"Start":"02:15.395 ","End":"02:18.980","Text":"let\u0027s call it c. This is just an existence."},{"Start":"02:18.980 ","End":"02:20.680","Text":"I don\u0027t know how to compute it or where it is,"},{"Start":"02:20.680 ","End":"02:23.215","Text":"but I know that it exists."},{"Start":"02:23.215 ","End":"02:27.050","Text":"At that point, the derivative of f is 0."},{"Start":"02:27.050 ","End":"02:28.595","Text":"Now, I sketch too,"},{"Start":"02:28.595 ","End":"02:32.095","Text":"because this is hard to visualize."},{"Start":"02:32.095 ","End":"02:34.160","Text":"Here are some axes,"},{"Start":"02:34.160 ","End":"02:37.235","Text":"and I\u0027ll do a sketch of a function."},{"Start":"02:37.235 ","End":"02:44.435","Text":"Let\u0027s say, something like this,"},{"Start":"02:44.435 ","End":"02:47.135","Text":"that this be our f of x,"},{"Start":"02:47.135 ","End":"02:48.980","Text":"y equals f of x,"},{"Start":"02:48.980 ","End":"02:53.375","Text":"and we\u0027ll let the points that I\u0027m interested in,"},{"Start":"02:53.375 ","End":"02:57.495","Text":"be this point, and this point."},{"Start":"02:57.495 ","End":"02:59.715","Text":"This will be a,"},{"Start":"02:59.715 ","End":"03:02.040","Text":"this will be b."},{"Start":"03:02.040 ","End":"03:07.340","Text":"We can see that f of a equals f of b."},{"Start":"03:07.340 ","End":"03:09.920","Text":"That\u0027s condition 3."},{"Start":"03:09.920 ","End":"03:16.140","Text":"The function between a and b is certainly continuous."},{"Start":"03:16.550 ","End":"03:22.285","Text":"We can sketch it without taking the pen off the paper, so to speak."},{"Start":"03:22.285 ","End":"03:24.860","Text":"It\u0027s also differentiable."},{"Start":"03:24.860 ","End":"03:26.780","Text":"It looks like it\u0027s differentiable everywhere."},{"Start":"03:26.780 ","End":"03:31.445","Text":"Differentiable means smooth without any sharp corners or anything like that."},{"Start":"03:31.445 ","End":"03:34.975","Text":"I\u0027m assuming it satisfies these 3 conditions."},{"Start":"03:34.975 ","End":"03:38.325","Text":"What this tells us is that,"},{"Start":"03:38.325 ","End":"03:44.415","Text":"there exists a point c somewhere between a and b,"},{"Start":"03:44.415 ","End":"03:46.470","Text":"where the derivative is 0."},{"Start":"03:46.470 ","End":"03:49.095","Text":"Let\u0027s look. Between a and b,"},{"Start":"03:49.095 ","End":"03:50.310","Text":"I\u0027m going from here to here,"},{"Start":"03:50.310 ","End":"03:52.400","Text":"and I can see that at this point,"},{"Start":"03:52.400 ","End":"03:56.945","Text":"the derivative is 0 because the tangent is horizontal."},{"Start":"03:56.945 ","End":"04:01.730","Text":"This point, if I take its x coordinate,"},{"Start":"04:01.730 ","End":"04:05.745","Text":"c is the c that is promised in the theorem."},{"Start":"04:05.745 ","End":"04:11.835","Text":"There actually could be more than 1 c in between a and b that satisfies this."},{"Start":"04:11.835 ","End":"04:13.730","Text":"The theorem guarantees there exists 1,"},{"Start":"04:13.730 ","End":"04:15.035","Text":"but there might be more."},{"Start":"04:15.035 ","End":"04:18.365","Text":"For example, suppose instead of taking b here,"},{"Start":"04:18.365 ","End":"04:20.840","Text":"I took it somewhere over here."},{"Start":"04:20.840 ","End":"04:23.615","Text":"I move the point b over here,"},{"Start":"04:23.615 ","End":"04:26.795","Text":"and f of a is still equal to f of b,"},{"Start":"04:26.795 ","End":"04:30.895","Text":"still continuous between a and b, undifferentiable."},{"Start":"04:30.895 ","End":"04:34.115","Text":"In this case, you can see that"},{"Start":"04:34.115 ","End":"04:39.770","Text":"the original c is still a possible c that satisfies this condition."},{"Start":"04:39.770 ","End":"04:43.325","Text":"But there\u0027s another 1 over here,"},{"Start":"04:43.325 ","End":"04:46.040","Text":"where f prime is 0."},{"Start":"04:46.040 ","End":"04:48.935","Text":"There\u0027s also a c_2, there could be any number of them."},{"Start":"04:48.935 ","End":"04:50.960","Text":"The theorem just says that there exists 1,"},{"Start":"04:50.960 ","End":"04:52.625","Text":"but there could be more than 1."},{"Start":"04:52.625 ","End":"04:59.730","Text":"There is an important corollary or conclusion from Rolle\u0027s theorem."},{"Start":"05:01.550 ","End":"05:04.500","Text":"This corollary states that,"},{"Start":"05:04.500 ","End":"05:12.955","Text":"if the function f of x cuts the x-axis at least twice,"},{"Start":"05:12.955 ","End":"05:21.360","Text":"then f prime of x that the derivative cuts the x-axis at least once."},{"Start":"05:23.680 ","End":"05:26.030","Text":"You might be wondering,"},{"Start":"05:26.030 ","End":"05:29.555","Text":"how on earth is this related to Rolle\u0027s theorem?"},{"Start":"05:29.555 ","End":"05:34.400","Text":"Let\u0027s just assume though that f satisfies the conditions as in Rolle\u0027s theorem,"},{"Start":"05:34.400 ","End":"05:36.650","Text":"but it\u0027s continuous and differentiable."},{"Start":"05:36.650 ","End":"05:38.180","Text":"I\u0027ll draw a sketch,"},{"Start":"05:38.180 ","End":"05:41.590","Text":"so you\u0027ll understand better what I\u0027m talking about."},{"Start":"05:41.590 ","End":"05:43.880","Text":"Here\u0027s a sketch."},{"Start":"05:43.880 ","End":"05:45.785","Text":"We have a function f of x,"},{"Start":"05:45.785 ","End":"05:49.655","Text":"and does indeed cut the x-axis twice."},{"Start":"05:49.655 ","End":"05:53.875","Text":"It cuts it here, and here."},{"Start":"05:53.875 ","End":"05:57.450","Text":"Let\u0027s call this point x equals a,"},{"Start":"05:57.450 ","End":"06:00.045","Text":"and here x equals b."},{"Start":"06:00.045 ","End":"06:02.090","Text":"1 thing\u0027s for sure,"},{"Start":"06:02.090 ","End":"06:08.995","Text":"it satisfies f of a equals f of b because it\u0027s 0 at both."},{"Start":"06:08.995 ","End":"06:12.490","Text":"We already assumed that we have continuity and differentiability."},{"Start":"06:12.490 ","End":"06:14.920","Text":"Now, we can apply Rolle\u0027s theorem."},{"Start":"06:14.920 ","End":"06:19.540","Text":"From Rolle\u0027s theorem, we conclude that somewhere between a and b,"},{"Start":"06:19.540 ","End":"06:21.450","Text":"there\u0027s a point c. Well,"},{"Start":"06:21.450 ","End":"06:23.455","Text":"here in this picture I can see it."},{"Start":"06:23.455 ","End":"06:26.710","Text":"The point c, where f prime is 0,"},{"Start":"06:26.710 ","End":"06:28.900","Text":"the derivative is 0 at this point,"},{"Start":"06:28.900 ","End":"06:35.540","Text":"because you can see the tangent is parallel to the x-axis, has a slope of 0."},{"Start":"06:35.540 ","End":"06:37.769","Text":"Now, we\u0027ve satisfied the conditions,"},{"Start":"06:37.769 ","End":"06:41.185","Text":"so we can say that the derivative cuts the x-axis."},{"Start":"06:41.185 ","End":"06:44.290","Text":"If I draw f prime on the same graph,"},{"Start":"06:44.290 ","End":"06:48.760","Text":"at this point, c will be 0 and I don\u0027t know what it looks like."},{"Start":"06:48.760 ","End":"06:53.225","Text":"Something that goes through this point here."},{"Start":"06:53.225 ","End":"06:55.730","Text":"Yeah, something like that. That\u0027s f prime of x."},{"Start":"06:55.730 ","End":"06:57.770","Text":"It will certainly, at the same point,"},{"Start":"06:57.770 ","End":"06:59.840","Text":"c cut the x-axis."},{"Start":"06:59.840 ","End":"07:03.800","Text":"This corollary is important and it\u0027s also useful."},{"Start":"07:03.800 ","End":"07:10.220","Text":"We can solve exercises with it and very soon I\u0027ll give you an example."},{"Start":"07:10.220 ","End":"07:17.135","Text":"I just would like to suggest that it\u0027s so important that you learn it by heart."},{"Start":"07:17.135 ","End":"07:19.925","Text":"There I\u0027ve highlighted it and I\u0027m going to repeat it."},{"Start":"07:19.925 ","End":"07:24.275","Text":"If the function f of x cuts the x-axis at least twice,"},{"Start":"07:24.275 ","End":"07:30.779","Text":"then the derivative function f prime of x cuts the x-axis at least once."},{"Start":"07:31.480 ","End":"07:35.600","Text":"In fact, the corollary is what\u0027s useful."},{"Start":"07:35.600 ","End":"07:45.680","Text":"The question it\u0027s useful for is to show that an equation has at most, 1 solution."},{"Start":"07:45.680 ","End":"07:51.785","Text":"The corollary is useful for showing that an equation f of x equals 0 has,"},{"Start":"07:51.785 ","End":"07:56.015","Text":"at most, I want to underline this at most,1 solution."},{"Start":"07:56.015 ","End":"07:57.515","Text":"Why do I underline it?"},{"Start":"07:57.515 ","End":"07:59.030","Text":"Because you\u0027re used to the opposite."},{"Start":"07:59.030 ","End":"08:04.030","Text":"For example, the intermediate value theorem"},{"Start":"08:04.030 ","End":"08:07.165","Text":"is useful for showing that it has at least 1 solution."},{"Start":"08:07.165 ","End":"08:09.820","Text":"Might remember, if f is positive somewhere,"},{"Start":"08:09.820 ","End":"08:12.040","Text":"and negative somewhere else and it\u0027s continuous,"},{"Start":"08:12.040 ","End":"08:15.275","Text":"then it has at least 1 solution."},{"Start":"08:15.275 ","End":"08:18.095","Text":"But here we\u0027re talking about at most,"},{"Start":"08:18.095 ","End":"08:21.920","Text":"I think it\u0027s about time we got to that example."},{"Start":"08:21.920 ","End":"08:25.790","Text":"I\u0027ll just give us some space."},{"Start":"08:25.790 ","End":"08:29.540","Text":"Here we come to the example exercise."},{"Start":"08:29.540 ","End":"08:36.260","Text":"Prove that the equation x^3 plus"},{"Start":"08:36.260 ","End":"08:44.400","Text":"4x plus 10 equals 0 has at most, 1 solution."},{"Start":"08:47.170 ","End":"08:50.960","Text":"Let\u0027s do the solution."},{"Start":"08:50.960 ","End":"09:00.620","Text":"First of all, notice that if I take the function f of x is equal to x^3 plus 4x plus 10."},{"Start":"09:00.620 ","End":"09:04.625","Text":"Then to say that it\u0027s equal to 0,"},{"Start":"09:04.625 ","End":"09:11.300","Text":"say that this is 0 means that f of x is 0 means that the function cuts the x-axis."},{"Start":"09:11.300 ","End":"09:16.775","Text":"This question is equivalent to showing that f of x"},{"Start":"09:16.775 ","End":"09:24.995","Text":"cuts the x-axis, at most, once."},{"Start":"09:24.995 ","End":"09:30.770","Text":"That\u0027s what the original request is equivalent to that of showing that this equation"},{"Start":"09:30.770 ","End":"09:33.500","Text":"has at most 1 solution will show that the function cuts"},{"Start":"09:33.500 ","End":"09:36.680","Text":"the x-axis at most 1, and that\u0027s equivalent."},{"Start":"09:36.680 ","End":"09:38.495","Text":"How do we do this?"},{"Start":"09:38.495 ","End":"09:42.230","Text":"Well, we\u0027re talking about cutting the x-axis."},{"Start":"09:42.230 ","End":"09:46.505","Text":"It looks like we could use this corollary which I copied here."},{"Start":"09:46.505 ","End":"09:51.830","Text":"What I\u0027m going to do is what is called in mathematics a proof by contradiction."},{"Start":"09:51.830 ","End":"09:54.920","Text":"If it\u0027s not true,"},{"Start":"09:54.920 ","End":"09:56.945","Text":"that it cuts at most once,"},{"Start":"09:56.945 ","End":"09:59.015","Text":"it means that it cuts more than once."},{"Start":"09:59.015 ","End":"10:00.710","Text":"That is at least twice."},{"Start":"10:00.710 ","End":"10:03.065","Text":"I\u0027ll call it proof by contradiction."},{"Start":"10:03.065 ","End":"10:07.970","Text":"We assume that what we\u0027re trying to prove is not true and we reach a contradiction."},{"Start":"10:07.970 ","End":"10:14.704","Text":"We assume that f cuts the x-axis at least twice."},{"Start":"10:14.704 ","End":"10:17.075","Text":"That\u0027s the opposite of at most once,"},{"Start":"10:17.075 ","End":"10:19.590","Text":"is at least twice."},{"Start":"10:19.780 ","End":"10:23.030","Text":"Then we\u0027ll reach a contradiction."},{"Start":"10:23.030 ","End":"10:29.165","Text":"By the corollary, if it cuts the x-axis at least twice,"},{"Start":"10:29.165 ","End":"10:35.360","Text":"then f prime cuts the x-axis at least once."},{"Start":"10:35.360 ","End":"10:37.925","Text":"I should have mentioned that, of course,"},{"Start":"10:37.925 ","End":"10:43.610","Text":"f being a polynomial is certainly continuous and differentiable everywhere."},{"Start":"10:43.610 ","End":"10:48.440","Text":"It satisfies the Rolle\u0027s theorem and the corollary also,"},{"Start":"10:48.440 ","End":"10:51.860","Text":"it should be mentioned that f is continuous and differentiable,"},{"Start":"10:51.860 ","End":"10:55.970","Text":"but f prime is not something abstract."},{"Start":"10:55.970 ","End":"10:57.050","Text":"I mean, I know what it is."},{"Start":"10:57.050 ","End":"11:00.635","Text":"I\u0027m given the formula so I can actually differentiate it."},{"Start":"11:00.635 ","End":"11:02.840","Text":"This means f prime is"},{"Start":"11:02.840 ","End":"11:10.730","Text":"3x^2 plus 4 cuts the x-axis but look at this function."},{"Start":"11:10.730 ","End":"11:13.910","Text":"I mean x squared is non-negative and so is"},{"Start":"11:13.910 ","End":"11:18.515","Text":"3x squared it\u0027s at least 0 and if I add 4 it\u0027s at least 4."},{"Start":"11:18.515 ","End":"11:20.120","Text":"If this is always at least 4,"},{"Start":"11:20.120 ","End":"11:22.820","Text":"there\u0027s no way it can cut the axis."},{"Start":"11:22.820 ","End":"11:25.595","Text":"Now, we say contradiction,"},{"Start":"11:25.595 ","End":"11:28.640","Text":"because this function does not cut the x-axis."},{"Start":"11:28.640 ","End":"11:31.205","Text":"Where did the contradiction come from?"},{"Start":"11:31.205 ","End":"11:37.490","Text":"The contradiction came from assuming that it cuts at least twice by negating this."},{"Start":"11:37.490 ","End":"11:40.010","Text":"If we negate this and get a contradiction,"},{"Start":"11:40.010 ","End":"11:41.570","Text":"it must mean that this is right,"},{"Start":"11:41.570 ","End":"11:44.045","Text":"because the alternative is a contradiction."},{"Start":"11:44.045 ","End":"11:46.369","Text":"That proves the example."},{"Start":"11:46.369 ","End":"11:51.540","Text":"I\u0027d like to show you an alternative solution that doesn\u0027t use Rolle\u0027s theorem."},{"Start":"11:53.530 ","End":"11:58.415","Text":"Typically, the alternative solution to this question is to"},{"Start":"11:58.415 ","End":"12:03.635","Text":"explore the intervals of increase, decrease, and extrema."},{"Start":"12:03.635 ","End":"12:08.850","Text":"In this case, we look at f prime of x."},{"Start":"12:08.890 ","End":"12:12.260","Text":"We\u0027ve already found that in the previous question."},{"Start":"12:12.260 ","End":"12:15.780","Text":"There it is 3x^2 plus 4."},{"Start":"12:15.940 ","End":"12:18.485","Text":"If we analyze this,"},{"Start":"12:18.485 ","End":"12:20.930","Text":"we find that this is never 0,"},{"Start":"12:20.930 ","End":"12:23.810","Text":"so there is only 1 interval and it\u0027s positive."},{"Start":"12:23.810 ","End":"12:28.230","Text":"So this is always positive."},{"Start":"12:28.870 ","End":"12:35.670","Text":"That means that f of x is always increasing."},{"Start":"12:40.750 ","End":"12:43.880","Text":"If a function is always increasing,"},{"Start":"12:43.880 ","End":"12:52.250","Text":"it can\u0027t cut the x-axis more than once because it\u0027s always going up and if it crosses it,"},{"Start":"12:52.250 ","End":"12:54.260","Text":"it can\u0027t ever cross it back."},{"Start":"12:54.260 ","End":"12:57.200","Text":"I mean, it might not cut at all."},{"Start":"12:57.200 ","End":"13:01.880","Text":"I\u0027ll show you, I\u0027ll make some little mini graphs here, make a couple of them."},{"Start":"13:01.880 ","End":"13:04.565","Text":"Let\u0027s take 1 and let\u0027s take another 1."},{"Start":"13:04.565 ","End":"13:08.420","Text":"There\u0027s really only a couple of possibilities."},{"Start":"13:08.420 ","End":"13:11.240","Text":"I mean, if it\u0027s constantly increasing,"},{"Start":"13:11.240 ","End":"13:16.490","Text":"it might increase and"},{"Start":"13:16.490 ","End":"13:21.515","Text":"cut once like this or it might be constantly increasing and not cut at all."},{"Start":"13:21.515 ","End":"13:25.445","Text":"It might be something like this."},{"Start":"13:25.445 ","End":"13:30.680","Text":"Asymptotically approach something less than 0 or 0."},{"Start":"13:30.680 ","End":"13:35.375","Text":"So f of x can\u0027t cross the x-axis twice."},{"Start":"13:35.375 ","End":"13:36.965","Text":"I\u0027m constantly going up,"},{"Start":"13:36.965 ","End":"13:38.630","Text":"I can\u0027t cross twice."},{"Start":"13:38.630 ","End":"13:42.440","Text":"The same thing would work also if it was constantly decreasing."},{"Start":"13:42.440 ","End":"13:46.925","Text":"It can\u0027t cross the x-axis twice,"},{"Start":"13:46.925 ","End":"13:50.510","Text":"which means that it cuts it at most once."},{"Start":"13:50.510 ","End":"13:57.470","Text":"We have at most once [inaudible] That basically proves it."},{"Start":"13:57.470 ","End":"14:02.670","Text":"Won\u0027t write the complete sentence and we are done."}],"ID":6290},{"Watched":false,"Name":"Example 1","Duration":"4m 54s","ChapterTopicVideoID":6278,"CourseChapterTopicPlaylistID":84682,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6278.jpeg","UploadDate":"2019-11-14T07:18:45.6500000","DurationForVideoObject":"PT4M54S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.080","Text":"Now, let\u0027s do some examples of the use of Rolle\u0027s theorem."},{"Start":"00:04.080 ","End":"00:05.100","Text":"Here\u0027s the first one."},{"Start":"00:05.100 ","End":"00:06.675","Text":"Well, really it\u0027s 2 in 1."},{"Start":"00:06.675 ","End":"00:09.540","Text":"We have to check if the given function f satisfies"},{"Start":"00:09.540 ","End":"00:12.810","Text":"the conditions of Rolle\u0027s theorem on the given interval,"},{"Start":"00:12.810 ","End":"00:15.525","Text":"and if so we have to find the values of c"},{"Start":"00:15.525 ","End":"00:18.705","Text":"that are promised us in the conclusion of Rolle\u0027s theorem."},{"Start":"00:18.705 ","End":"00:21.075","Text":"I\u0027m not going to repeat Rolle\u0027s theorem,"},{"Start":"00:21.075 ","End":"00:23.295","Text":"but I\u0027ll remind you as we go along,"},{"Start":"00:23.295 ","End":"00:25.200","Text":"we have to check 3 things."},{"Start":"00:25.200 ","End":"00:29.760","Text":"Here\u0027s our function f of x defined on the interval from 0 to 2."},{"Start":"00:29.760 ","End":"00:33.975","Text":"We have to check that it\u0027s continuous on this interval."},{"Start":"00:33.975 ","End":"00:40.000","Text":"It\u0027s differentiable on the open interval and the value of f at the end points is equal."},{"Start":"00:40.000 ","End":"00:41.885","Text":"Now, this function is a polynomial."},{"Start":"00:41.885 ","End":"00:44.705","Text":"It\u0027s continuous and differentiable everywhere,"},{"Start":"00:44.705 ","End":"00:48.380","Text":"so there\u0027s no need to check specifically for this interval."},{"Start":"00:48.380 ","End":"00:50.270","Text":"The only thing we have to check really,"},{"Start":"00:50.270 ","End":"00:53.980","Text":"is that the value at 0 and the value of 2 are equal,"},{"Start":"00:53.980 ","End":"00:56.265","Text":"so f of 0,"},{"Start":"00:56.265 ","End":"01:00.410","Text":"if I put in 0 these terms are all 0, this is 0."},{"Start":"01:00.410 ","End":"01:03.305","Text":"If I check what happens at x equals 2,"},{"Start":"01:03.305 ","End":"01:08.895","Text":"I get 8 minus 12 plus 4 is also 0."},{"Start":"01:08.895 ","End":"01:10.685","Text":"The answer is yes,"},{"Start":"01:10.685 ","End":"01:12.350","Text":"because the endpoints are equal,"},{"Start":"01:12.350 ","End":"01:16.310","Text":"and let\u0027s write it down that we\u0027ve checked that it\u0027s continuous and it\u0027s"},{"Start":"01:16.310 ","End":"01:21.335","Text":"differentiable as with the first 2 conditions and this is the third condition."},{"Start":"01:21.335 ","End":"01:25.460","Text":"We are promised some value of c between 0 and 2,"},{"Start":"01:25.460 ","End":"01:32.240","Text":"where f prime of c is equal to 0 for something between 0 and 2."},{"Start":"01:32.240 ","End":"01:33.725","Text":"Let\u0027s see if we can find it."},{"Start":"01:33.725 ","End":"01:38.545","Text":"What we have to do is differentiate f and find an equation."},{"Start":"01:38.545 ","End":"01:47.130","Text":"f prime of x is equal to 3x squared minus 6x plus 2."},{"Start":"01:47.130 ","End":"01:51.315","Text":"Now, we want f prime of c equals 0,"},{"Start":"01:51.315 ","End":"01:58.400","Text":"which means that 3c squared minus 6c plus 2 equals 0."},{"Start":"01:58.400 ","End":"02:06.860","Text":"What we get is c equals 3 plus or minus the square root of 3/3."},{"Start":"02:06.860 ","End":"02:16.460","Text":"Now, approximately this one comes out to be about 1.6 under the minus about 0.4."},{"Start":"02:16.460 ","End":"02:18.350","Text":"This is for the plus, this is for the minus,"},{"Start":"02:18.350 ","End":"02:23.120","Text":"and this is approximately enough to see that it\u0027s between 0 and 2."},{"Start":"02:23.120 ","End":"02:25.525","Text":"Those are the 2 answers."},{"Start":"02:25.525 ","End":"02:27.590","Text":"That\u0027s part A."},{"Start":"02:27.590 ","End":"02:29.420","Text":"Now, for part B,"},{"Start":"02:29.420 ","End":"02:30.950","Text":"this time it\u0027s not a polynomial,"},{"Start":"02:30.950 ","End":"02:32.575","Text":"it\u0027s a rational function."},{"Start":"02:32.575 ","End":"02:36.730","Text":"The rational function is well behaved except where the denominator is 0."},{"Start":"02:36.730 ","End":"02:39.185","Text":"Denominator is 0 when x equals 2,"},{"Start":"02:39.185 ","End":"02:41.270","Text":"and we\u0027re looking at minus 1 to 1,"},{"Start":"02:41.270 ","End":"02:42.665","Text":"and so there\u0027s no problems here."},{"Start":"02:42.665 ","End":"02:44.930","Text":"Therefore this will also be continuous and"},{"Start":"02:44.930 ","End":"02:48.155","Text":"differentiable everywhere except where it\u0027s not defined."},{"Start":"02:48.155 ","End":"02:50.330","Text":"Continuous? Yes."},{"Start":"02:50.330 ","End":"02:52.860","Text":"Differentiable? Yes."},{"Start":"02:52.860 ","End":"02:54.740","Text":"Now, the final condition"},{"Start":"02:54.740 ","End":"02:57.665","Text":"that the value of f at both ends is the same."},{"Start":"02:57.665 ","End":"03:01.850","Text":"I need to check f of minus 1 and f of 1"},{"Start":"03:01.850 ","End":"03:04.640","Text":"and hope that it comes out the same or not."},{"Start":"03:04.640 ","End":"03:06.935","Text":"When x is minus 1,"},{"Start":"03:06.935 ","End":"03:11.715","Text":"minus 1 squared minus 1 is 0 so I don\u0027t need to check anymore that\u0027s 0."},{"Start":"03:11.715 ","End":"03:14.000","Text":"Again, 1 squared minus 1 is 0"},{"Start":"03:14.000 ","End":"03:15.410","Text":"so this also comes out 0"},{"Start":"03:15.410 ","End":"03:17.170","Text":"and these are equal."},{"Start":"03:17.170 ","End":"03:20.890","Text":"Yes, we\u0027ve got 3 yeses for the conditions, therefore,"},{"Start":"03:20.890 ","End":"03:22.130","Text":"we\u0027ve got the conclusion that"},{"Start":"03:22.130 ","End":"03:26.150","Text":"we have some value of c between minus 1 and 1,"},{"Start":"03:26.150 ","End":"03:29.715","Text":"such that f prime of c equals 0."},{"Start":"03:29.715 ","End":"03:32.480","Text":"We\u0027re asked to find all the possible values of c."},{"Start":"03:32.480 ","End":"03:35.900","Text":"We\u0027re going to differentiate and set to 0,"},{"Start":"03:35.900 ","End":"03:38.800","Text":"f prime of x is equal to,"},{"Start":"03:38.800 ","End":"03:43.310","Text":"using the quotient rule I\u0027ve got the derivative of the numerator times"},{"Start":"03:43.310 ","End":"03:48.970","Text":"the denominator minus the numerator times derivative of denominator,"},{"Start":"03:48.970 ","End":"03:54.485","Text":"that equals 2x squared minus x squared is x squared"},{"Start":"03:54.485 ","End":"04:00.530","Text":"minus 4x plus 1 over x minus 2 squared."},{"Start":"04:00.530 ","End":"04:05.255","Text":"Now, we want f prime of c to equal 0."},{"Start":"04:05.255 ","End":"04:08.240","Text":"We have x equals c, it should be 0."},{"Start":"04:08.240 ","End":"04:13.715","Text":"We get that c squared minus 4c plus 1 equals 0."},{"Start":"04:13.715 ","End":"04:16.490","Text":"I just ignored the denominator because if a fraction is 0,"},{"Start":"04:16.490 ","End":"04:18.185","Text":"its numerator is 0."},{"Start":"04:18.185 ","End":"04:24.995","Text":"From here, I get that c equals 2 plus or minus square root of 3."},{"Start":"04:24.995 ","End":"04:27.530","Text":"But we have to check which of the,"},{"Start":"04:27.530 ","End":"04:30.440","Text":"that is between minus 1 and 1."},{"Start":"04:30.440 ","End":"04:33.244","Text":"I have to make sure that c is in this interval."},{"Start":"04:33.244 ","End":"04:36.105","Text":"Now square root of 3 is about 1.7."},{"Start":"04:36.105 ","End":"04:39.860","Text":"2 plus 1.7 is certainly not in this interval,"},{"Start":"04:39.860 ","End":"04:44.210","Text":"but 2 minus 1.7 is about 0.3 is in the interval."},{"Start":"04:44.210 ","End":"04:48.560","Text":"The answer is just the 2 minus the square root of 3,"},{"Start":"04:48.560 ","End":"04:50.065","Text":"that\u0027s the only one."},{"Start":"04:50.065 ","End":"04:52.880","Text":"We are done with this exercise."},{"Start":"04:52.880 ","End":"04:55.290","Text":"Let\u0027s go on to the next."}],"ID":6291},{"Watched":false,"Name":"Example 2","Duration":"3m 25s","ChapterTopicVideoID":6279,"CourseChapterTopicPlaylistID":84682,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6279.jpeg","UploadDate":"2019-11-14T07:18:49.5930000","DurationForVideoObject":"PT3M25S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.205","Text":"This is example 2."},{"Start":"00:02.205 ","End":"00:03.870","Text":"We\u0027re given a function of x,"},{"Start":"00:03.870 ","End":"00:06.134","Text":"1 over x minus 3 squared."},{"Start":"00:06.134 ","End":"00:10.140","Text":"We\u0027re asked to show that f of 1 equals f of 5,"},{"Start":"00:10.140 ","End":"00:13.950","Text":"but that there is no value of c between 1 and 5,"},{"Start":"00:13.950 ","End":"00:15.914","Text":"where the derivative is 0."},{"Start":"00:15.914 ","End":"00:18.900","Text":"Let\u0027s leave that last mysterious question to the end,"},{"Start":"00:18.900 ","End":"00:27.045","Text":"so f of 1 is 1 over 1 minus 3 squared which is 1/4,"},{"Start":"00:27.045 ","End":"00:32.870","Text":"f of 5 is 1 over 5 minus 3 squared,"},{"Start":"00:32.870 ","End":"00:34.400","Text":"which is a 1/4."},{"Start":"00:34.400 ","End":"00:37.880","Text":"F of 1 equals f of 5."},{"Start":"00:37.880 ","End":"00:41.300","Text":"The next bit about this C between 1 and 5,"},{"Start":"00:41.300 ","End":"00:46.850","Text":"first have to find what f prime is so f prime of x,"},{"Start":"00:46.850 ","End":"00:49.745","Text":"this is x minus 3 to the minus 2."},{"Start":"00:49.745 ","End":"00:52.460","Text":"Using exponents to the power of minus 2,"},{"Start":"00:52.460 ","End":"00:55.959","Text":"so the derivative is going to be minus 2,"},{"Start":"00:55.959 ","End":"00:58.350","Text":"x minus 3 to the minus 3,"},{"Start":"00:58.350 ","End":"01:01.385","Text":"and the minus 3, I\u0027m going to put it back in the denominator."},{"Start":"01:01.385 ","End":"01:05.720","Text":"We would like f prime of c to be 0. Can it be 0?"},{"Start":"01:05.720 ","End":"01:10.805","Text":"There\u0027s no point in even writing the equation because the numerator is never 0,"},{"Start":"01:10.805 ","End":"01:14.645","Text":"minus 2 is never 0 so this is never 0."},{"Start":"01:14.645 ","End":"01:18.185","Text":"Particular, it\u0027s not 0 between 1 and 5."},{"Start":"01:18.185 ","End":"01:20.090","Text":"Now, what is this they\u0027re asking us,"},{"Start":"01:20.090 ","End":"01:22.370","Text":"does this contradict Rolle\u0027s theorem?"},{"Start":"01:22.370 ","End":"01:25.399","Text":"Well, let\u0027s remember what Rolle\u0027s theorem says."},{"Start":"01:25.399 ","End":"01:29.705","Text":"Talking about the function f on the interval from 1-5."},{"Start":"01:29.705 ","End":"01:31.070","Text":"What Rolle\u0027s theorem says,"},{"Start":"01:31.070 ","End":"01:33.575","Text":"basically if f of 1 equals f of 5,"},{"Start":"01:33.575 ","End":"01:37.400","Text":"then there\u0027s supposed to be a value c in between 1 and 5,"},{"Start":"01:37.400 ","End":"01:39.670","Text":"where the derivative is 0."},{"Start":"01:39.670 ","End":"01:43.080","Text":"What\u0027s happened here is, was Rolle mistaken."},{"Start":"01:43.080 ","End":"01:45.120","Text":"Well, certainly not because there wasn\u0027t"},{"Start":"01:45.120 ","End":"01:48.110","Text":"just the 1 condition that the endpoints are equal."},{"Start":"01:48.110 ","End":"01:49.775","Text":"I mean the functions equal there,"},{"Start":"01:49.775 ","End":"01:51.335","Text":"there were 3 conditions."},{"Start":"01:51.335 ","End":"01:54.025","Text":"There was a condition of continuity."},{"Start":"01:54.025 ","End":"01:55.445","Text":"That\u0027s 1 condition."},{"Start":"01:55.445 ","End":"01:58.595","Text":"There was a condition 2 of differentiability"},{"Start":"01:58.595 ","End":"02:02.195","Text":"and there was a section 3 about the equality at both ends."},{"Start":"02:02.195 ","End":"02:04.835","Text":"We proved the equality part,"},{"Start":"02:04.835 ","End":"02:07.280","Text":"but the continuity doesn\u0027t hold."},{"Start":"02:07.280 ","End":"02:09.095","Text":"I mean, look at this function."},{"Start":"02:09.095 ","End":"02:11.525","Text":"When x is 3, it\u0027s not defined."},{"Start":"02:11.525 ","End":"02:16.460","Text":"But not defined at some point it\u0027s certainly not going to be continuous so this is no,"},{"Start":"02:16.460 ","End":"02:19.220","Text":"you don\u0027t even have to check this. Well, it\u0027s also no."},{"Start":"02:19.220 ","End":"02:21.230","Text":"It\u0027s not defined, it\u0027s not differentiable,"},{"Start":"02:21.230 ","End":"02:25.325","Text":"but it doesn\u0027t meet all 3 conditions and so it doesn\u0027t contradict."},{"Start":"02:25.325 ","End":"02:28.630","Text":"I\u0027ll draw a sketch just to give you a better idea."},{"Start":"02:28.630 ","End":"02:31.120","Text":"Normally when we imagine Rolle,"},{"Start":"02:31.120 ","End":"02:34.705","Text":"we imagine something like a sum function."},{"Start":"02:34.705 ","End":"02:36.810","Text":"We\u0027re given 2 points,"},{"Start":"02:36.810 ","End":"02:41.485","Text":"say here and here, an a and a b,"},{"Start":"02:41.485 ","End":"02:46.570","Text":"where the function is equal and then we say between here there\u0027s going to be"},{"Start":"02:46.570 ","End":"02:52.630","Text":"a point where the derivative is 0 and that point we\u0027ll call c and everything\u0027s fine."},{"Start":"02:52.630 ","End":"02:57.580","Text":"What happens here is that with our function there\u0027s a hole in the middle,"},{"Start":"02:57.580 ","End":"03:00.600","Text":"if c is 3, and a and b are 1 and 5,"},{"Start":"03:00.600 ","End":"03:02.790","Text":"what we have is a hole here."},{"Start":"03:02.790 ","End":"03:07.919","Text":"There\u0027s a hole, there is no point c in the domain and so it\u0027s ruins everything."},{"Start":"03:07.919 ","End":"03:10.940","Text":"It\u0027s important when you do Rolle\u0027s theorem to really check"},{"Start":"03:10.940 ","End":"03:13.925","Text":"these 3 conditions because if we don\u0027t have all the conditions,"},{"Start":"03:13.925 ","End":"03:17.660","Text":"people often skip the continuity or differentiability."},{"Start":"03:17.660 ","End":"03:20.090","Text":"If you do that, you\u0027re not guaranteed the conclusions"},{"Start":"03:20.090 ","End":"03:22.655","Text":"of Rolle\u0027s theorem, so no contradiction."},{"Start":"03:22.655 ","End":"03:24.080","Text":"We just weren\u0027t careful."},{"Start":"03:24.080 ","End":"03:26.710","Text":"Done."}],"ID":6292},{"Watched":false,"Name":"Example 3","Duration":"4m 30s","ChapterTopicVideoID":6280,"CourseChapterTopicPlaylistID":84682,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6280.jpeg","UploadDate":"2019-11-14T07:18:54.4730000","DurationForVideoObject":"PT4M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.420","Text":"In this exercise, we have to prove that the equation,"},{"Start":"00:03.420 ","End":"00:08.085","Text":"x squared plus x cubed plus 5x equals 1 has exactly 1 solution,"},{"Start":"00:08.085 ","End":"00:10.800","Text":"and we\u0027re given a hint, show, first of all,"},{"Start":"00:10.800 ","End":"00:16.185","Text":"that it has at least one solution and then that it has at most one solution."},{"Start":"00:16.185 ","End":"00:18.045","Text":"If we prove the hint, then we\u0027re done."},{"Start":"00:18.045 ","End":"00:21.810","Text":"Because if a number is at least one and at most one,"},{"Start":"00:21.810 ","End":"00:23.490","Text":"then it\u0027s exactly one."},{"Start":"00:23.490 ","End":"00:25.950","Text":"Let\u0027s start with at least one solution."},{"Start":"00:25.950 ","End":"00:30.130","Text":"This question is usually solved with the intermediate value theorem."},{"Start":"00:30.130 ","End":"00:39.065","Text":"Let\u0027s define f of x is equal to x squared plus x cubed plus 5x minus 1."},{"Start":"00:39.065 ","End":"00:42.290","Text":"The original equation we were asked to solve becomes"},{"Start":"00:42.290 ","End":"00:46.730","Text":"exactly equivalent to saying that f of x equals 0."},{"Start":"00:46.730 ","End":"00:49.400","Text":"That\u0027s what this equation is equivalent to."},{"Start":"00:49.400 ","End":"00:51.450","Text":"What we\u0027re really asking,"},{"Start":"00:51.450 ","End":"00:59.000","Text":"it can be rephrased as f of x intersects the x axis exactly once,"},{"Start":"00:59.000 ","End":"01:02.195","Text":"and then we\u0027re going to continue as here, the exactly,"},{"Start":"01:02.195 ","End":"01:04.400","Text":"we\u0027re going to convert into first of all,"},{"Start":"01:04.400 ","End":"01:08.660","Text":"showing at least once and then at most once."},{"Start":"01:08.660 ","End":"01:12.440","Text":"The at least once is done with the intermediate value theorem."},{"Start":"01:12.440 ","End":"01:14.690","Text":"Let\u0027s just keep plugging in some values of x"},{"Start":"01:14.690 ","End":"01:17.135","Text":"until we get one positive and one negative."},{"Start":"01:17.135 ","End":"01:19.774","Text":"Let\u0027s try f of 0."},{"Start":"01:19.774 ","End":"01:21.530","Text":"That\u0027s easiest to compute."},{"Start":"01:21.530 ","End":"01:23.165","Text":"That\u0027s minus 1."},{"Start":"01:23.165 ","End":"01:25.970","Text":"Now, I need a positive f of x."},{"Start":"01:25.970 ","End":"01:30.020","Text":"Let\u0027s try f of 1, f of 1 is equal to c,"},{"Start":"01:30.020 ","End":"01:35.460","Text":"1 plus 1 plus 5 is 7, minus 1 is 6."},{"Start":"01:35.460 ","End":"01:36.740","Text":"We\u0027re good."},{"Start":"01:36.740 ","End":"01:40.039","Text":"This is negative, this is less than 0."},{"Start":"01:40.039 ","End":"01:41.870","Text":"This is bigger than 0."},{"Start":"01:41.870 ","End":"01:44.375","Text":"So we have at least one solution."},{"Start":"01:44.375 ","End":"01:47.385","Text":"We know there exists a value of c,"},{"Start":"01:47.385 ","End":"01:49.140","Text":"so that the c equals 0."},{"Start":"01:49.140 ","End":"01:51.710","Text":"That guarantees that there\u0027s at least one,"},{"Start":"01:51.710 ","End":"01:54.080","Text":"and now, how about at most once?"},{"Start":"01:54.080 ","End":"01:57.145","Text":"That\u0027s where we use the Corollary to Rolle\u0027s theorem."},{"Start":"01:57.145 ","End":"02:00.045","Text":"We\u0027ll prove by contradiction."},{"Start":"02:00.045 ","End":"02:06.845","Text":"We have to show that f of x intersects the x-axis at most once."},{"Start":"02:06.845 ","End":"02:12.065","Text":"Now what we\u0027ll do is use what is called proof by contradiction."},{"Start":"02:12.065 ","End":"02:16.580","Text":"It\u0027s a method of proof where we assume that what we\u0027re trying to prove isn\u0027t so,"},{"Start":"02:16.580 ","End":"02:19.350","Text":"and we reach a contradiction and so we reject that,"},{"Start":"02:19.350 ","End":"02:22.160","Text":"and then we\u0027re back to the original hypothesis."},{"Start":"02:22.160 ","End":"02:28.025","Text":"Proof by contradiction means that I\u0027ll assume that this thing is false,"},{"Start":"02:28.025 ","End":"02:31.760","Text":"that f does not intersect the x-axis at most once,"},{"Start":"02:31.760 ","End":"02:33.470","Text":"and I\u0027ll reach a contradiction."},{"Start":"02:33.470 ","End":"02:34.700","Text":"If we reach a contradiction,"},{"Start":"02:34.700 ","End":"02:36.500","Text":"we have to reject the alternative,"},{"Start":"02:36.500 ","End":"02:38.420","Text":"which means accepting this."},{"Start":"02:38.420 ","End":"02:43.175","Text":"What is the opposite of f of x intersects the axis at most once."},{"Start":"02:43.175 ","End":"02:45.910","Text":"The opposite is that it intersects at least twice."},{"Start":"02:45.910 ","End":"02:50.015","Text":"Let\u0027s say, f of x intersects twice or more."},{"Start":"02:50.015 ","End":"02:56.090","Text":"By Rolle\u0027s Corollary, it implies that f prime of x intersects at least once,"},{"Start":"02:56.090 ","End":"03:00.260","Text":"and if it intersects once at some point c,"},{"Start":"03:00.260 ","End":"03:04.895","Text":"then we have f prime of c equals 0."},{"Start":"03:04.895 ","End":"03:07.069","Text":"Now, we can compute f prime."},{"Start":"03:07.069 ","End":"03:13.265","Text":"If f of x is x squared plus x cubed plus 5x minus 1,"},{"Start":"03:13.265 ","End":"03:21.875","Text":"then f prime of x in general is equal to 2x plus 3x squared plus 5."},{"Start":"03:21.875 ","End":"03:24.365","Text":"Now, let\u0027s check when this is 0."},{"Start":"03:24.365 ","End":"03:27.619","Text":"If f prime of c equals 0,"},{"Start":"03:27.619 ","End":"03:29.645","Text":"then by the quadratic formula,"},{"Start":"03:29.645 ","End":"03:31.460","Text":"I\u0027ll just rewrite it in a different order"},{"Start":"03:31.460 ","End":"03:33.290","Text":"because we\u0027re not used to this order,"},{"Start":"03:33.290 ","End":"03:38.060","Text":"then that means that c is minus 2 plus or"},{"Start":"03:38.060 ","End":"03:44.270","Text":"minus the square root of minus 56 all over 6."},{"Start":"03:44.270 ","End":"03:49.460","Text":"This minus 56 means that there is no answer, there is no such c,"},{"Start":"03:49.460 ","End":"03:54.245","Text":"and so that is a contradiction because the Corollary said there would be such a c,"},{"Start":"03:54.245 ","End":"03:55.930","Text":"but there is no such c,"},{"Start":"03:55.930 ","End":"03:58.655","Text":"and where did this contradiction come from?"},{"Start":"03:58.655 ","End":"04:02.615","Text":"The contradiction, assuming that f could intersect twice,"},{"Start":"04:02.615 ","End":"04:04.295","Text":"which is the negation of this."},{"Start":"04:04.295 ","End":"04:06.500","Text":"So if this can\u0027t be false,"},{"Start":"04:06.500 ","End":"04:07.955","Text":"because if you assume it\u0027s false,"},{"Start":"04:07.955 ","End":"04:10.630","Text":"it reach a contradiction, so it must be true."},{"Start":"04:10.630 ","End":"04:13.890","Text":"Because this statement was proved false,"},{"Start":"04:13.890 ","End":"04:15.530","Text":"this statement must be true."},{"Start":"04:15.530 ","End":"04:17.300","Text":"One of these 2 has to be true."},{"Start":"04:17.300 ","End":"04:19.220","Text":"That means that, yes,"},{"Start":"04:19.220 ","End":"04:21.470","Text":"it intersects most ones."},{"Start":"04:21.470 ","End":"04:23.900","Text":"We\u0027ve already done the at least once,"},{"Start":"04:23.900 ","End":"04:25.745","Text":"and now we\u0027ve just proved this,"},{"Start":"04:25.745 ","End":"04:27.030","Text":"and we have this and this,"},{"Start":"04:27.030 ","End":"04:28.760","Text":"so we have the exactly."},{"Start":"04:28.760 ","End":"04:31.170","Text":"We\u0027re done."}],"ID":6293}],"Thumbnail":null,"ID":84682},{"Name":"Summarizing Questions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Question 1","Duration":"2m 30s","ChapterTopicVideoID":11456,"CourseChapterTopicPlaylistID":257165,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/11456.jpeg","UploadDate":"2018-02-05T16:28:54.1930000","DurationForVideoObject":"PT2M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:05.175","Text":"In this exercise, we are going to be using the intermediate value theorem."},{"Start":"00:05.175 ","End":"00:10.800","Text":"We\u0027re given a function f that is continuous on the interval from 0 to 8."},{"Start":"00:10.800 ","End":"00:19.935","Text":"We\u0027re also told that the only value in this interval where f is 0 is at x equals 5."},{"Start":"00:19.935 ","End":"00:30.240","Text":"F of 5 is 0, but no other x in this interval is 0 of f. Now we have to prove that if f"},{"Start":"00:30.240 ","End":"00:35.100","Text":"of 3 is bigger than"},{"Start":"00:35.100 ","End":"00:42.995","Text":"0 then so is f of 4 and there\u0027s the hint to use that theorem."},{"Start":"00:42.995 ","End":"00:46.515","Text":"The prove is prove by contradiction."},{"Start":"00:46.515 ","End":"00:50.870","Text":"We suppose that what we\u0027re going to prove is not true."},{"Start":"00:50.870 ","End":"00:57.020","Text":"Suppose that f of 4 is less than 0."},{"Start":"00:57.020 ","End":"01:00.110","Text":"Why didn\u0027t I write less than or equal to?"},{"Start":"01:00.110 ","End":"01:03.980","Text":"Because I know that the only place where"},{"Start":"01:03.980 ","End":"01:09.100","Text":"0 is 5 so it\u0027s not bigger than it\u0027s actually smaller than 0."},{"Start":"01:09.100 ","End":"01:18.425","Text":"But f of 3 is bigger than 0."},{"Start":"01:18.425 ","End":"01:21.949","Text":"Now look we have at 4 the function is negative,"},{"Start":"01:21.949 ","End":"01:24.690","Text":"at 3 it\u0027s positive."},{"Start":"01:28.190 ","End":"01:33.585","Text":"By the intermediate value theorem,"},{"Start":"01:33.585 ","End":"01:38.450","Text":"we know that f of c is equal to 0 for"},{"Start":"01:38.450 ","End":"01:46.950","Text":"some c between 3 and 4."},{"Start":"01:46.950 ","End":"01:49.305","Text":"Since I used x here,"},{"Start":"01:49.305 ","End":"01:51.950","Text":"let\u0027s change the c to an x, it doesn\u0027t matter."},{"Start":"01:51.950 ","End":"01:57.475","Text":"Another f of x equals 0 for some x between 3 and 4."},{"Start":"01:57.475 ","End":"01:59.910","Text":"Now, can x be 5?"},{"Start":"01:59.910 ","End":"02:02.505","Text":"X can\u0027t be 5."},{"Start":"02:02.505 ","End":"02:10.345","Text":"That means there is another place where f of x is 0 and that\u0027s a contradiction."},{"Start":"02:10.345 ","End":"02:16.550","Text":"This contradiction contradicts that the only place in 0 is at"},{"Start":"02:16.550 ","End":"02:22.400","Text":"x equals 5 and that contradiction means that the supposition is false,"},{"Start":"02:22.400 ","End":"02:27.850","Text":"that f of 4 is bigger than 0."},{"Start":"02:27.850 ","End":"02:30.640","Text":"Okay, that\u0027s it."}],"ID":11820},{"Watched":false,"Name":"Question 2","Duration":"1m 33s","ChapterTopicVideoID":11457,"CourseChapterTopicPlaylistID":257165,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/11457.jpeg","UploadDate":"2018-02-05T16:29:04.7430000","DurationForVideoObject":"PT1M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:06.465","Text":"In this exercise, we have a function f graphed below."},{"Start":"00:06.465 ","End":"00:11.970","Text":"It\u0027s continuous on the closed interval from a to b"},{"Start":"00:11.970 ","End":"00:17.430","Text":"and it\u0027s differentiable on the open interval from a to b."},{"Start":"00:17.430 ","End":"00:22.780","Text":"All this means that it satisfies the conditions of the mean value theorem."},{"Start":"00:23.780 ","End":"00:28.305","Text":"Only one of the numbers that I\u0027ve labeled,"},{"Start":"00:28.305 ","End":"00:31.620","Text":"the x value, C_1, C_2, C_3,"},{"Start":"00:31.620 ","End":"00:38.895","Text":"and C_4 satisfy the conclusion of the mean value theorem on this interval."},{"Start":"00:38.895 ","End":"00:40.675","Text":"Which one is it?"},{"Start":"00:40.675 ","End":"00:48.920","Text":"Remember the mean value theorem guarantees that there is a point in the middle"},{"Start":"00:48.920 ","End":"00:58.055","Text":"where the tangent line is parallel to the line that joins f of a and f of b."},{"Start":"00:58.055 ","End":"01:00.545","Text":"Well, the green line here."},{"Start":"01:00.545 ","End":"01:08.505","Text":"Let\u0027s draw some of the tangent lines to the 4 points here."},{"Start":"01:08.505 ","End":"01:15.440","Text":"Let\u0027s decide which one of these looks like it could be parallel to this one."},{"Start":"01:15.440 ","End":"01:22.790","Text":"I think there\u0027s no doubt that the only one it could be is this one,"},{"Start":"01:22.790 ","End":"01:28.595","Text":"so the answer is that it is C_4."},{"Start":"01:28.595 ","End":"01:30.950","Text":"That\u0027s all there is to it."},{"Start":"01:30.950 ","End":"01:33.720","Text":"So we are done."}],"ID":11821},{"Watched":false,"Name":"Question 3","Duration":"1m 49s","ChapterTopicVideoID":11458,"CourseChapterTopicPlaylistID":257165,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/11458.jpeg","UploadDate":"2018-02-05T16:29:19.2030000","DurationForVideoObject":"PT1M49S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.370","Text":"In this exercise, we have to show that the cubic polynomial,"},{"Start":"00:05.370 ","End":"00:09.735","Text":"it\u0027s written here, has a root in the interval 0, 1."},{"Start":"00:09.735 ","End":"00:13.095","Text":"We\u0027re given a hint to use the Intermediate Value Theorem."},{"Start":"00:13.095 ","End":"00:16.870","Text":"A polynomial is also a function."},{"Start":"00:17.330 ","End":"00:24.284","Text":"As a function, it\u0027s certainly continuous everywhere and in the interval,"},{"Start":"00:24.284 ","End":"00:25.470","Text":"and because it\u0027s continuous,"},{"Start":"00:25.470 ","End":"00:28.245","Text":"we can use the Intermediate Value Theorem."},{"Start":"00:28.245 ","End":"00:34.070","Text":"Now the obvious thing to do is to check the value at the end points"},{"Start":"00:34.070 ","End":"00:36.770","Text":"and hope that we have a positive and a negative."},{"Start":"00:36.770 ","End":"00:38.480","Text":"Well, let\u0027s see."},{"Start":"00:38.480 ","End":"00:51.265","Text":"P of 0 equals just 2, and P of 1 is equal to 1 minus 2 minus 3 plus 2."},{"Start":"00:51.265 ","End":"00:54.945","Text":"That comes out to minus 2."},{"Start":"00:54.945 ","End":"01:00.780","Text":"This is positive and this is negative,"},{"Start":"01:00.780 ","End":"01:03.920","Text":"and because it\u0027s continuous,"},{"Start":"01:03.920 ","End":"01:10.040","Text":"any value between 2 and minus 2 is attained somewhere in the interval,"},{"Start":"01:10.040 ","End":"01:17.670","Text":"in particular 0, because 0 is between minus 2 and 2."},{"Start":"01:17.670 ","End":"01:19.560","Text":"There is some point c,"},{"Start":"01:19.560 ","End":"01:23.240","Text":"such that P of c is equal to 0."},{"Start":"01:23.240 ","End":"01:25.010","Text":"We can\u0027t say what c is."},{"Start":"01:25.010 ","End":"01:27.500","Text":"Well, in this case, we could probably compute it,"},{"Start":"01:27.500 ","End":"01:30.544","Text":"but just to show that it exists,"},{"Start":"01:30.544 ","End":"01:35.400","Text":"so c would be the root of P."},{"Start":"01:35.400 ","End":"01:43.450","Text":"By the theorem also c is between 0 and 1."},{"Start":"01:43.790 ","End":"01:45.260","Text":"That\u0027s it."},{"Start":"01:45.260 ","End":"01:50.310","Text":"Just an elementary use of the Intermediate Value Theorem."}],"ID":11822},{"Watched":false,"Name":"Question 4","Duration":"1m 47s","ChapterTopicVideoID":11459,"CourseChapterTopicPlaylistID":257165,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/11459.jpeg","UploadDate":"2018-02-05T16:29:27.5000000","DurationForVideoObject":"PT1M47S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.590","Text":"In this exercise, we\u0027re given some facts about a function f."},{"Start":"00:04.590 ","End":"00:09.300","Text":"We\u0027re told it\u0027s continuous and differentiable on the interval from 0 to 5."},{"Start":"00:09.300 ","End":"00:11.040","Text":"That f of 0,"},{"Start":"00:11.040 ","End":"00:14.360","Text":"that\u0027s f at the left end point is minus 2,"},{"Start":"00:14.360 ","End":"00:18.929","Text":"and we\u0027re also told that f prime of x less than or equal to 2,"},{"Start":"00:18.929 ","End":"00:21.255","Text":"meaning for all x in the interval."},{"Start":"00:21.255 ","End":"00:28.060","Text":"The question is, what\u0027s the largest possible value for f at the other end point, f of 5."},{"Start":"00:28.060 ","End":"00:30.770","Text":"Because f is continuous and differentiable,"},{"Start":"00:30.770 ","End":"00:33.620","Text":"we can use the mean value theorem."},{"Start":"00:33.620 ","End":"00:37.715","Text":"This gives us that f of 5,"},{"Start":"00:37.715 ","End":"00:40.700","Text":"minus f of 0,"},{"Start":"00:40.700 ","End":"00:44.135","Text":"over 5 minus 0,"},{"Start":"00:44.135 ","End":"00:48.320","Text":"equals f prime of c. We don\u0027t know what c is,"},{"Start":"00:48.320 ","End":"00:54.810","Text":"but there exists at least 1 c in the interval from 0-5."},{"Start":"00:55.870 ","End":"00:58.820","Text":"Now, let\u0027s use some of the facts were told, well,"},{"Start":"00:58.820 ","End":"01:01.970","Text":"we know that f of 0 is minus 2,"},{"Start":"01:01.970 ","End":"01:05.540","Text":"so f of 5 minus,"},{"Start":"01:05.540 ","End":"01:10.755","Text":"minus 2 is plus 2 over 5."},{"Start":"01:10.755 ","End":"01:13.709","Text":"Now, f prime of c,"},{"Start":"01:13.709 ","End":"01:15.695","Text":"c is 1 of the x\u0027s,"},{"Start":"01:15.695 ","End":"01:17.510","Text":"so it\u0027s less than or equal to 2."},{"Start":"01:17.510 ","End":"01:21.064","Text":"We get that this is less than or equal to 2."},{"Start":"01:21.064 ","End":"01:24.805","Text":"Multiply both sides by 5,"},{"Start":"01:24.805 ","End":"01:30.495","Text":"f of 5 plus 2 less than or equal to 10,"},{"Start":"01:30.495 ","End":"01:32.640","Text":"5 is positive so we can multiply,"},{"Start":"01:32.640 ","End":"01:34.560","Text":"and now subtract 2,"},{"Start":"01:34.560 ","End":"01:39.720","Text":"from both sides, and we\u0027ve got f of 5 less than or equal to 8."},{"Start":"01:39.720 ","End":"01:44.960","Text":"8 is the maximum possible value for f of 5."},{"Start":"01:44.960 ","End":"01:47.580","Text":"We are done."}],"ID":11823},{"Watched":false,"Name":"Question 5","Duration":"4m 33s","ChapterTopicVideoID":11460,"CourseChapterTopicPlaylistID":257165,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/11460.jpeg","UploadDate":"2018-02-05T16:29:47.8430000","DurationForVideoObject":"PT4M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.420","Text":"In this exercise, we\u0027re given this function f of x cubic polynomial,"},{"Start":"00:06.420 ","End":"00:11.145","Text":"and we have to show that it has exactly 1 real root."},{"Start":"00:11.145 ","End":"00:14.430","Text":"Given a hint that we\u0027re going to use 2 theorems,"},{"Start":"00:14.430 ","End":"00:18.060","Text":"the intermediate value theorem and Rolle\u0027s theorem."},{"Start":"00:18.060 ","End":"00:20.610","Text":"In case you haven\u0027t heard of Rolle\u0027s theorem is"},{"Start":"00:20.610 ","End":"00:22.950","Text":"a special case of the intermediate value theorem."},{"Start":"00:22.950 ","End":"00:27.090","Text":"I\u0027ll mention that when we get to it."},{"Start":"00:27.090 ","End":"00:32.475","Text":"The first step is to show that it has a root, at least one."},{"Start":"00:32.475 ","End":"00:35.760","Text":"We\u0027re going to use this intermediate value theorem."},{"Start":"00:35.760 ","End":"00:39.115","Text":"What we\u0027re going to do is find the value of x where"},{"Start":"00:39.115 ","End":"00:43.370","Text":"the function is positive and another one where it\u0027s negative and then it will follow."},{"Start":"00:43.370 ","End":"00:48.575","Text":"Let\u0027s try an easy substitution f of 0."},{"Start":"00:48.575 ","End":"00:54.045","Text":"All these are 0\u0027s, so that\u0027s minus 4 and that is negative."},{"Start":"00:54.045 ","End":"00:57.640","Text":"We want f of something to be positive."},{"Start":"00:57.640 ","End":"01:05.795","Text":"Certainly, there\u0027s going to be 1 because this function goes to infinity on the right."},{"Start":"01:05.795 ","End":"01:08.570","Text":"We\u0027ll just substitute x equals 1, x equals 2."},{"Start":"01:08.570 ","End":"01:09.905","Text":"Eventually we\u0027ll get there."},{"Start":"01:09.905 ","End":"01:12.625","Text":"Let\u0027s try f of 1."},{"Start":"01:12.625 ","End":"01:21.910","Text":"That is 1 minus 6 plus 15 minus 4."},{"Start":"01:21.920 ","End":"01:25.185","Text":"Yeah, that\u0027s already positive."},{"Start":"01:25.185 ","End":"01:31.005","Text":"Good. We know that somewhere between 0 and 1,"},{"Start":"01:31.005 ","End":"01:40.800","Text":"there is some value c such that f of c is equal to 0."},{"Start":"01:40.800 ","End":"01:43.385","Text":"We have at least 1 root."},{"Start":"01:43.385 ","End":"01:46.340","Text":"Now I\u0027m going to go to the second part,"},{"Start":"01:46.340 ","End":"01:48.830","Text":"which is to show that it can\u0027t have more than 1."},{"Start":"01:48.830 ","End":"01:50.750","Text":"Suppose it has 2 roots."},{"Start":"01:50.750 ","End":"01:53.435","Text":"Let\u0027s call them c_1 and c_2."},{"Start":"01:53.435 ","End":"02:01.405","Text":"F of c_1 is 0 and f of c_2 is 0 and will reach a contradiction."},{"Start":"02:01.405 ","End":"02:08.055","Text":"By Rolle\u0027s Theorem, there is another c, it\u0027s called a c_3,"},{"Start":"02:08.055 ","End":"02:13.440","Text":"such that c_3 is between c_1 and c_2,"},{"Start":"02:13.440 ","End":"02:19.115","Text":"and the derivative at c_3 is 0."},{"Start":"02:19.115 ","End":"02:21.260","Text":"If you don\u0027t know about Rolle\u0027s Theorem,"},{"Start":"02:21.260 ","End":"02:28.715","Text":"we use the intermediate value theorem and we would say f prime of c_3 equals"},{"Start":"02:28.715 ","End":"02:37.145","Text":"f of c_2 minus f of c_1 over c_2 minus c_1."},{"Start":"02:37.145 ","End":"02:39.665","Text":"F of c_2 is 0,"},{"Start":"02:39.665 ","End":"02:41.974","Text":"f of c_1 is 0."},{"Start":"02:41.974 ","End":"02:44.935","Text":"This thing comes out 0."},{"Start":"02:44.935 ","End":"02:46.880","Text":"C_2 is not equal to c_1,"},{"Start":"02:46.880 ","End":"02:49.175","Text":"of course, as we\u0027re assuming that they\u0027re different."},{"Start":"02:49.175 ","End":"02:51.020","Text":"Again, you\u0027d get it."},{"Start":"02:51.020 ","End":"02:56.080","Text":"You could use the intermediate value theorem instead of Rolle\u0027s Theorem."},{"Start":"02:56.600 ","End":"03:01.695","Text":"Now, what is the problem with f prime of c_3 equaling 0?"},{"Start":"03:01.695 ","End":"03:07.675","Text":"I\u0027ll show you. Let\u0027s figure out what f prime of x is equal to."},{"Start":"03:07.675 ","End":"03:16.890","Text":"It\u0027s 3x squared minus 12x plus 15."},{"Start":"03:16.890 ","End":"03:22.790","Text":"I\u0027m going to show that this doesn\u0027t have any real roots at all."},{"Start":"03:22.790 ","End":"03:27.055","Text":"Suppose this is equal to 0."},{"Start":"03:27.055 ","End":"03:30.560","Text":"Let\u0027s just divide by 3, make it easier."},{"Start":"03:30.560 ","End":"03:36.110","Text":"X squared minus 4x plus 5 equals 0."},{"Start":"03:36.110 ","End":"03:38.420","Text":"To check that it has no roots,"},{"Start":"03:38.420 ","End":"03:42.110","Text":"all we have to do is check the discriminant,"},{"Start":"03:42.110 ","End":"03:44.540","Text":"b squared minus 4ac,"},{"Start":"03:44.540 ","End":"03:46.595","Text":"the bit that\u0027s under the square root,"},{"Start":"03:46.595 ","End":"03:55.455","Text":"which is 4 squared minus 4 times 1 times 5,"},{"Start":"03:55.455 ","End":"03:58.050","Text":"it\u0027s 16 minus 20,"},{"Start":"03:58.050 ","End":"03:59.845","Text":"and is minus 4."},{"Start":"03:59.845 ","End":"04:01.940","Text":"In any event it\u0027s negative,"},{"Start":"04:01.940 ","End":"04:04.550","Text":"so has no square root."},{"Start":"04:04.550 ","End":"04:06.619","Text":"This is the discriminant,"},{"Start":"04:06.619 ","End":"04:08.885","Text":"by the way, sometimes called Delta."},{"Start":"04:08.885 ","End":"04:10.670","Text":"When this is negative,"},{"Start":"04:10.670 ","End":"04:19.010","Text":"the quadratic equation has no roots and so it\u0027s impossible for any real number,"},{"Start":"04:19.010 ","End":"04:22.565","Text":"even c_3 to be a root of it."},{"Start":"04:22.565 ","End":"04:26.210","Text":"This contradiction shows that it can\u0027t have 2 roots,"},{"Start":"04:26.210 ","End":"04:28.720","Text":"but it does have at least one root."},{"Start":"04:28.720 ","End":"04:31.410","Text":"It has exactly one real root."},{"Start":"04:31.410 ","End":"04:34.150","Text":"We are done."}],"ID":11824}],"Thumbnail":null,"ID":257165},{"Name":"Advanced Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"1m 1s","ChapterTopicVideoID":25344,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25344.jpeg","UploadDate":"2021-04-01T16:21:26.6870000","DurationForVideoObject":"PT1M1S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.120","Text":"In this exercise, we were asked whether there exists"},{"Start":"00:03.120 ","End":"00:07.050","Text":"a differentiable function f from the interval 0,"},{"Start":"00:07.050 ","End":"00:11.250","Text":"2 to the reals which satisfies f of 0 is minus 1,"},{"Start":"00:11.250 ","End":"00:18.840","Text":"f of 2 is 4, and f prime of x less than or equal to 2 for all x in the interval 0, 2."},{"Start":"00:18.840 ","End":"00:25.090","Text":"The answer is no and we\u0027ll prove this by contradiction."},{"Start":"00:25.090 ","End":"00:28.325","Text":"Suppose that such an f exists."},{"Start":"00:28.325 ","End":"00:32.675","Text":"By the mean value theorem on 0, 2,"},{"Start":"00:32.675 ","End":"00:37.235","Text":"we get that there exists some c strictly between 0 and 2,"},{"Start":"00:37.235 ","End":"00:45.754","Text":"such that f prime of c is f of 2 minus f of 0 over 2 minus 0."},{"Start":"00:45.754 ","End":"00:48.395","Text":"This comes out to be 2 1/2."},{"Start":"00:48.395 ","End":"00:54.550","Text":"But we\u0027re also told that f prime of x is less than or equal to 2."},{"Start":"00:54.550 ","End":"01:01.860","Text":"This is a contradiction because 2 1/2 is obviously not less than 2 so we are done."}],"ID":26161},{"Watched":false,"Name":"Exercise 2","Duration":"1m 24s","ChapterTopicVideoID":25345,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25345.jpeg","UploadDate":"2021-04-01T16:21:56.4900000","DurationForVideoObject":"PT1M24S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.030","Text":"In this exercise, f is differentiable on the unit interval 0, 1,"},{"Start":"00:07.030 ","End":"00:13.330","Text":"and the derivative has absolute value less than 1 on all this interval,"},{"Start":"00:13.330 ","End":"00:18.580","Text":"we have to show that there exists most 1 c in this interval,"},{"Start":"00:18.580 ","End":"00:23.400","Text":"such that f of c equals c. We\u0027re going to do"},{"Start":"00:23.400 ","End":"00:29.130","Text":"a proof by contradiction and we\u0027ll suppose that there are 2 such c,"},{"Start":"00:29.130 ","End":"00:33.045","Text":"f of c_1 is c_1 and f of c_2 is c_2."},{"Start":"00:33.045 ","End":"00:36.890","Text":"I guess I should have written that c_1 is not equal to c_2."},{"Start":"00:36.890 ","End":"00:39.654","Text":"Now by the mean value theorem,"},{"Start":"00:39.654 ","End":"00:45.290","Text":"there exists some c naught in the open interval c_1,"},{"Start":"00:45.290 ","End":"00:54.900","Text":"c_2, such that f of c_2 minus f of c_1 is f prime of c naught c_2 minus c_1."},{"Start":"00:54.900 ","End":"00:58.770","Text":"But f of c_2 is c_2 and f of c_1 is c_1."},{"Start":"00:58.770 ","End":"01:01.230","Text":"So what we get is this."},{"Start":"01:01.230 ","End":"01:04.490","Text":"Then just divide both sides by c_2"},{"Start":"01:04.490 ","End":"01:10.010","Text":"minus c_1 and we get that f prime of c naught is exactly equal to 1."},{"Start":"01:10.010 ","End":"01:16.100","Text":"This is a contradiction because for all x including c,"},{"Start":"01:16.100 ","End":"01:20.150","Text":"f prime of x is going to be less than 1, not equal 1."},{"Start":"01:20.150 ","End":"01:25.020","Text":"So this contradiction completes our proof. We\u0027re done."}],"ID":26162},{"Watched":false,"Name":"Exercise 3","Duration":"1m 37s","ChapterTopicVideoID":25346,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25346.jpeg","UploadDate":"2021-04-01T16:22:57.3230000","DurationForVideoObject":"PT1M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.305","Text":"In this exercise, f is real-valued is differentiable."},{"Start":"00:04.305 ","End":"00:06.600","Text":"For some constant Alpha,"},{"Start":"00:06.600 ","End":"00:11.505","Text":"the absolute value of f prime is less than or equal to Alpha,"},{"Start":"00:11.505 ","End":"00:18.090","Text":"less than 1 for all x. Alpha is between 0 and 1 but strictly less than 1."},{"Start":"00:18.090 ","End":"00:21.700","Text":"Let a_1 be some real number,"},{"Start":"00:21.700 ","End":"00:25.865","Text":"and we\u0027ll use it to generate a sequence a_n recursively."},{"Start":"00:25.865 ","End":"00:30.065","Text":"Each successive a is f of the previous a."},{"Start":"00:30.065 ","End":"00:34.130","Text":"We have to show that the sequence a_n converges."},{"Start":"00:34.130 ","End":"00:38.195","Text":"We\u0027re going to show that the sequence satisfies the contractive condition."},{"Start":"00:38.195 ","End":"00:43.085","Text":"Absolute value of a_n plus 2 minus a_n plus 1 equals,"},{"Start":"00:43.085 ","End":"00:45.275","Text":"using the recursive definition,"},{"Start":"00:45.275 ","End":"00:51.130","Text":"applying it twice, once with n and once with n plus 1, we get this."},{"Start":"00:51.130 ","End":"00:54.410","Text":"Applying the mean value theorem,"},{"Start":"00:54.410 ","End":"01:02.065","Text":"this is equal to f prime of c times absolute value of a_n plus 1 minus a_n."},{"Start":"01:02.065 ","End":"01:06.445","Text":"I didn\u0027t write it, but c is between a_n and a_n plus 1."},{"Start":"01:06.445 ","End":"01:08.885","Text":"I didn\u0027t write it because we don\u0027t need that fact."},{"Start":"01:08.885 ","End":"01:16.025","Text":"In any case, the absolute value of f prime of c is less than or equal to Alpha."},{"Start":"01:16.025 ","End":"01:18.335","Text":"What we have is less than or equal to Alpha,"},{"Start":"01:18.335 ","End":"01:21.140","Text":"absolute value of a_n plus 1 minus a_n."},{"Start":"01:21.140 ","End":"01:23.015","Text":"If we look at this and this,"},{"Start":"01:23.015 ","End":"01:26.635","Text":"this is exactly the contractive condition."},{"Start":"01:26.635 ","End":"01:31.130","Text":"From this, we conclude that the sequence a_n satisfies"},{"Start":"01:31.130 ","End":"01:38.370","Text":"the Cauchy criterion and hence it converges. We are done."}],"ID":26163},{"Watched":false,"Name":"Exercise 4","Duration":"1m 53s","ChapterTopicVideoID":25347,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25347.jpeg","UploadDate":"2021-04-01T16:23:52.1170000","DurationForVideoObject":"PT1M53S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.420","Text":"In this exercise, f is defined on the interval 0,"},{"Start":"00:03.420 ","End":"00:05.955","Text":"1 and is twice differentiable."},{"Start":"00:05.955 ","End":"00:10.860","Text":"Now we suppose that the line joining the points 0,"},{"Start":"00:10.860 ","End":"00:13.200","Text":"f 0 and 1,"},{"Start":"00:13.200 ","End":"00:17.820","Text":"f of 1 intersect the graph at a point a,"},{"Start":"00:17.820 ","End":"00:22.170","Text":"f of a, where a is between 0 and 1."},{"Start":"00:22.170 ","End":"00:26.820","Text":"We have to show that there exists an x naught in the interval 0,"},{"Start":"00:26.820 ","End":"00:30.329","Text":"1, where the second derivative is 0."},{"Start":"00:30.329 ","End":"00:35.385","Text":"Now by the mean value theorem on the interval from 0 to a,"},{"Start":"00:35.385 ","End":"00:44.050","Text":"we conclude that there is some b between 0 and a such that geometrically means that"},{"Start":"00:44.050 ","End":"00:47.740","Text":"the slope of the tangent is equal to the slope of"},{"Start":"00:47.740 ","End":"00:54.210","Text":"the chord and this is what it is written out, so that\u0027s b."},{"Start":"00:54.210 ","End":"00:59.260","Text":"Then doing the same thing on the interval from a to 1 we"},{"Start":"00:59.260 ","End":"01:05.079","Text":"get a point c between a and 1 that c,"},{"Start":"01:05.079 ","End":"01:13.345","Text":"where again, the slope of the tangent equals the slope of this line here."},{"Start":"01:13.345 ","End":"01:16.030","Text":"Now these 2 quantities are equal,"},{"Start":"01:16.030 ","End":"01:18.220","Text":"1 of them is the slope of this chord,"},{"Start":"01:18.220 ","End":"01:19.660","Text":"the other is the slope of this chord,"},{"Start":"01:19.660 ","End":"01:23.229","Text":"but they\u0027re all the same because of the slope of this line."},{"Start":"01:23.229 ","End":"01:28.460","Text":"In each case, it\u0027s rise over run for this line."},{"Start":"01:28.460 ","End":"01:30.424","Text":"So these 2 are the same,"},{"Start":"01:30.424 ","End":"01:34.279","Text":"which means that these 2 have got to be the same."},{"Start":"01:34.279 ","End":"01:36.229","Text":"Now by Rolle\u0027s theorem,"},{"Start":"01:36.229 ","End":"01:39.155","Text":"if f prime here equals f prime here,"},{"Start":"01:39.155 ","End":"01:42.050","Text":"then at some point in between them,"},{"Start":"01:42.050 ","End":"01:43.970","Text":"the derivative of f prime,"},{"Start":"01:43.970 ","End":"01:46.325","Text":"in other words, f double prime is 0."},{"Start":"01:46.325 ","End":"01:51.740","Text":"There is some x naught it\u0027s not on the diagram where f double-prime of x naught is 0,"},{"Start":"01:51.740 ","End":"01:54.300","Text":"and we are done."}],"ID":26164},{"Watched":false,"Name":"Exercise 5","Duration":"1m 58s","ChapterTopicVideoID":25348,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25348.jpeg","UploadDate":"2021-04-01T16:24:53.9900000","DurationForVideoObject":"PT1M58S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.655","Text":"In this exercise, f is continuous on the closed unit interval,"},{"Start":"00:05.655 ","End":"00:09.780","Text":"and it\u0027s differentiable on the open unit interval."},{"Start":"00:09.780 ","End":"00:16.605","Text":"We\u0027re given that the limit on the right of f prime is L,"},{"Start":"00:16.605 ","End":"00:19.515","Text":"as x goes to 0."},{"Start":"00:19.515 ","End":"00:23.280","Text":"We have to show that the right derivative,"},{"Start":"00:23.280 ","End":"00:24.600","Text":"that\u0027s what the plus here means."},{"Start":"00:24.600 ","End":"00:30.180","Text":"The right derivative of f at 0 exists and is equal to L,"},{"Start":"00:30.180 ","End":"00:37.800","Text":"same L. Now what this means is that the limit as h goes to 0 from"},{"Start":"00:37.800 ","End":"00:42.830","Text":"the right of f of h minus f of 0 over h minus"},{"Start":"00:42.830 ","End":"00:48.930","Text":"0 exists and is equal to L. That\u0027s what we need to show."},{"Start":"00:49.240 ","End":"00:54.935","Text":"By the mean value theorem if h is bigger than 0,"},{"Start":"00:54.935 ","End":"01:01.700","Text":"then this quotient is equal to f prime of some c,"},{"Start":"01:01.700 ","End":"01:04.775","Text":"which depends on h, so we called it c_h,"},{"Start":"01:04.775 ","End":"01:10.350","Text":"where c_h is between 0 and h. In a way,"},{"Start":"01:10.350 ","End":"01:14.705","Text":"c is a function of h. Note that by the sandwich theorem,"},{"Start":"01:14.705 ","End":"01:17.360","Text":"when h goes to 0,"},{"Start":"01:17.360 ","End":"01:20.030","Text":"then c_h goes to 0."},{"Start":"01:20.030 ","End":"01:26.450","Text":"It\u0027s clear. What we get is that the limit as h goes to 0 from the right"},{"Start":"01:26.450 ","End":"01:33.720","Text":"of f of h minus f of 0 over h is the limit as h goes to 0 of f prime of c_h."},{"Start":"01:34.070 ","End":"01:36.810","Text":"Now, like we said,"},{"Start":"01:36.810 ","End":"01:39.060","Text":"when h goes to 0,"},{"Start":"01:39.060 ","End":"01:42.300","Text":"c_h also goes to 0."},{"Start":"01:42.300 ","End":"01:49.130","Text":"This is what we have and this we\u0027re given is equal to L because you could think of c_h"},{"Start":"01:49.130 ","End":"01:56.655","Text":"is like the x here so we get this L. That\u0027s what we needed to prove."},{"Start":"01:56.655 ","End":"01:59.380","Text":"That concludes this clip."}],"ID":26165},{"Watched":false,"Name":"Exercise 6","Duration":"2m 54s","ChapterTopicVideoID":25349,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25349.jpeg","UploadDate":"2021-04-01T16:26:10.4730000","DurationForVideoObject":"PT2M54S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.339","Text":"In this exercise, f is a differentiable function on the closed unit interval,"},{"Start":"00:05.339 ","End":"00:08.580","Text":"and f of 0 is 0."},{"Start":"00:08.580 ","End":"00:13.260","Text":"We have an inequality that the absolute value of the derivative is"},{"Start":"00:13.260 ","End":"00:18.150","Text":"less than or equal to the absolute value of the function on our interval."},{"Start":"00:18.150 ","End":"00:25.290","Text":"What we have to show is that f is 0 on all of the interval."},{"Start":"00:25.290 ","End":"00:29.910","Text":"Let\u0027s take any x in 0, 1."},{"Start":"00:29.910 ","End":"00:32.200","Text":"We apply the mean value theorem,"},{"Start":"00:32.200 ","End":"00:38.494","Text":"and we get that there exists some x_1 between 0 and x,"},{"Start":"00:38.494 ","End":"00:45.020","Text":"such that f of x minus f of 0 is f prime of x_1 times x minus 0."},{"Start":"00:45.020 ","End":"00:48.410","Text":"Since f of 0 is 0,"},{"Start":"00:48.410 ","End":"00:55.010","Text":"this comes out to be just that f of x equals f prime of x_1 times x."},{"Start":"00:55.010 ","End":"00:59.570","Text":"Now take the absolute value and use"},{"Start":"00:59.570 ","End":"01:02.120","Text":"the fact that the absolute value of f prime of"},{"Start":"01:02.120 ","End":"01:05.720","Text":"x is less than or equal to the absolute value of f of x."},{"Start":"01:05.720 ","End":"01:08.500","Text":"We get this."},{"Start":"01:08.500 ","End":"01:14.055","Text":"Next, we\u0027re going to repeat the procedure but with x_1 in place of x."},{"Start":"01:14.055 ","End":"01:19.950","Text":"We get an x_2 between 0 and x_1 such that,"},{"Start":"01:19.950 ","End":"01:24.430","Text":"basically, what we wrote here only with x_2 in place of x_1,"},{"Start":"01:24.430 ","End":"01:26.545","Text":"and x_1 in place of x."},{"Start":"01:26.545 ","End":"01:32.610","Text":"If we replace f of x_1 that was here by this,"},{"Start":"01:32.610 ","End":"01:36.940","Text":"and we also use the fact that x_1 is less than x,"},{"Start":"01:36.940 ","End":"01:42.485","Text":"then this becomes less than or equal to x squared times f of x_2."},{"Start":"01:42.485 ","End":"01:45.210","Text":"Now, we can continue this way getting x_3,"},{"Start":"01:45.210 ","End":"01:46.470","Text":"x_4, and so on."},{"Start":"01:46.470 ","End":"01:48.745","Text":"In other words, inductively or recursively,"},{"Start":"01:48.745 ","End":"01:53.630","Text":"we get a sequence x_n in the interval 0, 1,"},{"Start":"01:53.630 ","End":"01:59.105","Text":"such that f of x is less than or equal to x to the power of n, f of x_n."},{"Start":"01:59.105 ","End":"02:02.630","Text":"Now, f is continuous on a closed interval,"},{"Start":"02:02.630 ","End":"02:04.340","Text":"and so it\u0027s bounded there."},{"Start":"02:04.340 ","End":"02:12.185","Text":"Let\u0027s say the bound is M. We can now re-evaluate this to say that for all n,"},{"Start":"02:12.185 ","End":"02:19.190","Text":"the absolute value of f of x is less than or equal to x to the power of n times M. Now"},{"Start":"02:19.190 ","End":"02:27.010","Text":"the sequence x to the n goes to 0 because x is less than 1 in absolute value."},{"Start":"02:27.010 ","End":"02:29.565","Text":"Since this is true for all n,"},{"Start":"02:29.565 ","End":"02:31.015","Text":"that the sandwich theorem,"},{"Start":"02:31.015 ","End":"02:33.185","Text":"f of x is 0."},{"Start":"02:33.185 ","End":"02:38.270","Text":"Since we have to know that f of x is 0 on the closed interval,"},{"Start":"02:38.270 ","End":"02:41.075","Text":"we still have to show that f of 1 is 0."},{"Start":"02:41.075 ","End":"02:43.485","Text":"Well, this follows from continuity,"},{"Start":"02:43.485 ","End":"02:46.425","Text":"since f is differentiable, it\u0027s continuous,"},{"Start":"02:46.425 ","End":"02:49.020","Text":"and if it\u0027s 0 everywhere except 1,"},{"Start":"02:49.020 ","End":"02:54.610","Text":"then it\u0027s also 0 at 1. We are done."}],"ID":26166},{"Watched":false,"Name":"Exercise 7","Duration":"1m 53s","ChapterTopicVideoID":25350,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25350.jpeg","UploadDate":"2021-04-01T16:26:56.7170000","DurationForVideoObject":"PT1M53S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.550","Text":"In this exercise, f is a function defined on the non-negatives,"},{"Start":"00:05.550 ","End":"00:08.940","Text":"a real-valued function, and it\u0027s continuous."},{"Start":"00:08.940 ","End":"00:11.009","Text":"Its value at 0 is 0."},{"Start":"00:11.009 ","End":"00:17.625","Text":"We also are given that f prime exists on the positive x,"},{"Start":"00:17.625 ","End":"00:22.545","Text":"and we\u0027re told that f prime is increasing on the positives."},{"Start":"00:22.545 ","End":"00:25.170","Text":"We have to show that this function g,"},{"Start":"00:25.170 ","End":"00:27.060","Text":"which is f of x over x,"},{"Start":"00:27.060 ","End":"00:30.075","Text":"is increasing on the positives."},{"Start":"00:30.075 ","End":"00:36.040","Text":"Because g is well-defined since x is positive and so not 0."},{"Start":"00:36.040 ","End":"00:40.520","Text":"The strategy will be to show that g prime"},{"Start":"00:40.520 ","End":"00:44.960","Text":"is non-negative and that will show that g is increasing."},{"Start":"00:44.960 ","End":"00:47.330","Text":"How do we compute g prime?"},{"Start":"00:47.330 ","End":"00:52.195","Text":"I\u0027ll just remind you of the quotient rule for derivatives."},{"Start":"00:52.195 ","End":"00:56.960","Text":"Using that, you can check that this is what we get for g prime."},{"Start":"00:56.960 ","End":"01:01.235","Text":"Cancel x top and bottom and we get this."},{"Start":"01:01.235 ","End":"01:05.950","Text":"Now want to evaluate this using MVT,"},{"Start":"01:05.950 ","End":"01:07.805","Text":"the mean value theorem."},{"Start":"01:07.805 ","End":"01:12.200","Text":"We get that f of x minus f of 0 over x minus 0 is"},{"Start":"01:12.200 ","End":"01:18.249","Text":"f prime of c_x where c_x is somewhere between 0 and x."},{"Start":"01:18.249 ","End":"01:23.735","Text":"We can replace and put f prime of c_x in place of this."},{"Start":"01:23.735 ","End":"01:28.735","Text":"Now, we\u0027re told that f prime is increasing."},{"Start":"01:28.735 ","End":"01:33.300","Text":"Notice that c_x is less than x."},{"Start":"01:33.300 ","End":"01:37.095","Text":"If I apply f prime to these 2 quantities,"},{"Start":"01:37.095 ","End":"01:40.060","Text":"we\u0027ll get a less than or equal to."},{"Start":"01:40.220 ","End":"01:44.420","Text":"That means that the numerator here is"},{"Start":"01:44.420 ","End":"01:47.930","Text":"bigger or equal to 0 and that\u0027s what we wanted to show,"},{"Start":"01:47.930 ","End":"01:53.820","Text":"because that shows us that the function g is increasing and we are done."}],"ID":26167},{"Watched":false,"Name":"Exercise 8","Duration":"2m 36s","ChapterTopicVideoID":25351,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25351.jpeg","UploadDate":"2021-04-01T16:27:56.6130000","DurationForVideoObject":"PT2M36S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.395","Text":"In this exercise, a is a non-negative number,"},{"Start":"00:04.395 ","End":"00:09.450","Text":"and we have function f from the interval a, b."},{"Start":"00:09.450 ","End":"00:11.940","Text":"B is strictly bigger than a,"},{"Start":"00:11.940 ","End":"00:15.345","Text":"real-valued and it\u0027s differentiable."},{"Start":"00:15.345 ","End":"00:20.115","Text":"We have to use Cauchy\u0027s mean value theorem to show that"},{"Start":"00:20.115 ","End":"00:25.080","Text":"there exist c_1 and c_2 in the open interval a,"},{"Start":"00:25.080 ","End":"00:29.535","Text":"b, such that this equality holds."},{"Start":"00:29.535 ","End":"00:32.550","Text":"In case you forgot Cauchy\u0027s mean value theorem,"},{"Start":"00:32.550 ","End":"00:34.755","Text":"this is basically what it says."},{"Start":"00:34.755 ","End":"00:39.900","Text":"We\u0027re going to use this twice with different functions,"},{"Start":"00:39.900 ","End":"00:46.014","Text":"g. The first g we\u0027ll try is g_1 of x equals x, the identity function."},{"Start":"00:46.014 ","End":"00:50.959","Text":"Notice that g_1 prime is 1,"},{"Start":"00:50.959 ","End":"00:59.895","Text":"so that there exists g_1 of b is b,"},{"Start":"00:59.895 ","End":"01:06.530","Text":"g_1 of a is a equals f prime of c_1 over g prime,"},{"Start":"01:06.530 ","End":"01:07.925","Text":"we said is 1."},{"Start":"01:07.925 ","End":"01:12.980","Text":"In fact, the Cauchy\u0027s mean value theorem reverts to"},{"Start":"01:12.980 ","End":"01:14.960","Text":"the regular mean value theorem when"},{"Start":"01:14.960 ","End":"01:20.180","Text":"the function g is the identity function g of x equals x."},{"Start":"01:20.180 ","End":"01:24.990","Text":"Now let\u0027s take g_2 of x to be x squared."},{"Start":"01:24.990 ","End":"01:27.530","Text":"Applying the Cauchy\u0027s mean value theorem,"},{"Start":"01:27.530 ","End":"01:34.430","Text":"we get that there exists c_2 such that f of b minus"},{"Start":"01:34.430 ","End":"01:37.910","Text":"f of a over b squared minus a squared is"},{"Start":"01:37.910 ","End":"01:42.155","Text":"f prime of c_2 and the derivative this time is 2x,"},{"Start":"01:42.155 ","End":"01:46.530","Text":"so when we apply that to c_2, we get 2c_2."},{"Start":"01:46.550 ","End":"01:52.755","Text":"I didn\u0027t write it but both c_1 and c_2 are between a and b,"},{"Start":"01:52.755 ","End":"01:55.930","Text":"so that satisfies this."},{"Start":"01:56.260 ","End":"02:02.470","Text":"Now let\u0027s simplify f of b minus f of a over b squared minus a squared,"},{"Start":"02:02.470 ","End":"02:06.380","Text":"the denominator factorizes as difference of squares."},{"Start":"02:06.380 ","End":"02:09.260","Text":"We get f of b minus f of a over b minus a,"},{"Start":"02:09.260 ","End":"02:11.915","Text":"and we still have a 1 over b plus a,"},{"Start":"02:11.915 ","End":"02:16.280","Text":"this bit is equal to f prime of c_1 from here."},{"Start":"02:16.280 ","End":"02:19.995","Text":"F prime of c_2 over 2c_2 just stays,"},{"Start":"02:19.995 ","End":"02:21.995","Text":"this is just writing it out."},{"Start":"02:21.995 ","End":"02:25.660","Text":"F prime of c_1 and the 1 over b plus a,"},{"Start":"02:25.660 ","End":"02:26.950","Text":"put it in the denominator,"},{"Start":"02:26.950 ","End":"02:30.155","Text":"change the order, so we get this."},{"Start":"02:30.155 ","End":"02:33.720","Text":"That is exactly what we had to show,"},{"Start":"02:33.720 ","End":"02:36.790","Text":"that\u0027s this here, so we\u0027re done."}],"ID":26168},{"Watched":false,"Name":"Exercise 9","Duration":"3m 12s","ChapterTopicVideoID":29500,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/29500.jpeg","UploadDate":"2022-07-21T15:30:17.1170000","DurationForVideoObject":"PT3M12S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.150","Text":"In this exercise, we have a real-valued function f on the reals,"},{"Start":"00:04.150 ","End":"00:09.885","Text":"and there\u0027s a point c where the second derivative of f exists."},{"Start":"00:09.885 ","End":"00:13.470","Text":"We have to show using L\u0027Hospital\u0027s rule that"},{"Start":"00:13.470 ","End":"00:18.600","Text":"this second derivative is given by this limit as here."},{"Start":"00:18.600 ","End":"00:20.759","Text":"There\u0027s another part of the question."},{"Start":"00:20.759 ","End":"00:25.080","Text":"Well, we have to give an example where the above limit exists,"},{"Start":"00:25.080 ","End":"00:27.165","Text":"this limit on the left,"},{"Start":"00:27.165 ","End":"00:31.035","Text":"but that f double prime of c does not exist."},{"Start":"00:31.035 ","End":"00:34.785","Text":"Now, the second derivative of f exists."},{"Start":"00:34.785 ","End":"00:38.100","Text":"So the first derivative exists in some neighborhood of"},{"Start":"00:38.100 ","End":"00:42.860","Text":"c. L\u0027hospital\u0027s rule says that to compute this limit,"},{"Start":"00:42.860 ","End":"00:49.650","Text":"we can differentiate top and bottom and if this limit exists, so does this."},{"Start":"00:49.650 ","End":"00:54.650","Text":"Let\u0027s differentiate, get f prime of c plus h,"},{"Start":"00:54.650 ","End":"00:57.200","Text":"and the inner derivative is 1."},{"Start":"00:57.200 ","End":"01:03.230","Text":"This is a constant as far as h goes that\u0027s 0 and here we get f prime of c"},{"Start":"01:03.230 ","End":"01:09.920","Text":"minus h and there\u0027s a minus here because the inner derivative is minus 1."},{"Start":"01:09.920 ","End":"01:12.215","Text":"Now we\u0027ll do a bit of algebra on this."},{"Start":"01:12.215 ","End":"01:16.475","Text":"The 2 here can be taken outside"},{"Start":"01:16.475 ","End":"01:25.310","Text":"the brackets and we can add and subtract f prime of c,"},{"Start":"01:25.310 ","End":"01:27.845","Text":"and that won\u0027t change anything."},{"Start":"01:27.845 ","End":"01:32.525","Text":"We\u0027ve now got this to be the sum of 2 limits."},{"Start":"01:32.525 ","End":"01:37.805","Text":"Now both these limits express the second derivative"},{"Start":"01:37.805 ","End":"01:44.105","Text":"of f at c. This one\u0027s on the right and this one\u0027s the derivative on the left."},{"Start":"01:44.105 ","End":"01:47.145","Text":"But it is differentiable, they are both the same."},{"Start":"01:47.145 ","End":"01:52.010","Text":"They\u0027re both equal to f double prime of c or the derivative of f prime."},{"Start":"01:52.010 ","End":"01:56.510","Text":"This comes out to be just f double prime of c,"},{"Start":"01:56.510 ","End":"01:58.280","Text":"which is what we wanted."},{"Start":"01:58.280 ","End":"02:00.140","Text":"That\u0027s the first part of the question."},{"Start":"02:00.140 ","End":"02:05.960","Text":"Now we have to give this example and the example will be the signum,"},{"Start":"02:05.960 ","End":"02:08.570","Text":"the sign function of x,"},{"Start":"02:08.570 ","End":"02:12.680","Text":"meaning 1 when x is positive minus 1 when x is negative,"},{"Start":"02:12.680 ","End":"02:16.355","Text":"and 0 when x is 0."},{"Start":"02:16.355 ","End":"02:20.555","Text":"Let\u0027s take our point c to be the point 0."},{"Start":"02:20.555 ","End":"02:25.815","Text":"Note that f is an odd function as this symmetry."},{"Start":"02:25.815 ","End":"02:30.214","Text":"Computing this limit with c equals 0,"},{"Start":"02:30.214 ","End":"02:32.020","Text":"this is what we get,"},{"Start":"02:32.020 ","End":"02:33.710","Text":"f of 0 is 0,"},{"Start":"02:33.710 ","End":"02:36.680","Text":"so we\u0027re just left with this."},{"Start":"02:36.680 ","End":"02:40.070","Text":"Since f is an odd function,"},{"Start":"02:40.070 ","End":"02:44.390","Text":"we can pull the minus here out in front and we get f of"},{"Start":"02:44.390 ","End":"02:49.785","Text":"h minus f of h and this is just 0,"},{"Start":"02:49.785 ","End":"02:53.840","Text":"and the limit of 0 is 0."},{"Start":"02:53.840 ","End":"02:59.254","Text":"So the limit exists and it\u0027s equal to 0."},{"Start":"02:59.254 ","End":"03:06.875","Text":"But f double prime 0 doesn\u0027t exist because f is not even continuous as 0."},{"Start":"03:06.875 ","End":"03:12.940","Text":"This is the required example and we are done."}],"ID":31115},{"Watched":false,"Name":"Exercise 10","Duration":"3m 1s","ChapterTopicVideoID":25337,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25337.jpeg","UploadDate":"2022-07-21T15:29:35.2470000","DurationForVideoObject":"PT3M1S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.630","Text":"In this exercise, the function f is differentiable on the closed interval a, b."},{"Start":"00:06.630 ","End":"00:13.420","Text":"We have to show that if the derivative is never 0 on that interval,"},{"Start":"00:13.420 ","End":"00:16.380","Text":"1 of 2 situations must occur."},{"Start":"00:16.380 ","End":"00:19.680","Text":"Either the derivative of f is always"},{"Start":"00:19.680 ","End":"00:24.510","Text":"non-negative or it\u0027s always non-positive on the interval."},{"Start":"00:24.510 ","End":"00:28.760","Text":"First step is to prove that f is 1 to 1."},{"Start":"00:28.760 ","End":"00:30.980","Text":"We\u0027ll do that by contradiction."},{"Start":"00:30.980 ","End":"00:33.875","Text":"Now, what does it mean for it not to be 1 to 1?"},{"Start":"00:33.875 ","End":"00:36.320","Text":"It means that there are 2 different points,"},{"Start":"00:36.320 ","End":"00:40.790","Text":"x1 and x2, but f of x1 is equal to f of x2."},{"Start":"00:40.790 ","End":"00:42.200","Text":"Well, in that case,"},{"Start":"00:42.200 ","End":"00:46.860","Text":"we can use Rolle\u0027s theorem and find c strictly between"},{"Start":"00:46.860 ","End":"00:52.895","Text":"x1 and x2 such that the derivative of f at c is 0."},{"Start":"00:52.895 ","End":"00:56.060","Text":"Now, that contradicts this here,"},{"Start":"00:56.060 ","End":"00:58.070","Text":"that f prime is never 0."},{"Start":"00:58.070 ","End":"01:01.340","Text":"So that contradiction proves that f is 1 to 1."},{"Start":"01:01.340 ","End":"01:04.205","Text":"Now we have 1 to 1 and continuous,"},{"Start":"01:04.205 ","End":"01:07.130","Text":"continuous falls from differentiable."},{"Start":"01:07.130 ","End":"01:12.335","Text":"By an earlier exercise on the intermediate value property,"},{"Start":"01:12.335 ","End":"01:17.525","Text":"we get that f is either strictly increasing or strictly decreasing."},{"Start":"01:17.525 ","End":"01:21.380","Text":"Now in case you forgotten that exercise,"},{"Start":"01:21.380 ","End":"01:27.350","Text":"here it is, and you could pause this and study it."},{"Start":"01:27.350 ","End":"01:30.960","Text":"Now back to our exercise."},{"Start":"01:32.030 ","End":"01:34.460","Text":"Let\u0027s split into 2 cases."},{"Start":"01:34.460 ","End":"01:39.395","Text":"First of all, take the case where f is strictly increasing."},{"Start":"01:39.395 ","End":"01:46.565","Text":"Then we\u0027ll show that this is what we get the first of the 2 possibilities."},{"Start":"01:46.565 ","End":"01:49.205","Text":"Now from the definition of the derivative,"},{"Start":"01:49.205 ","End":"01:52.355","Text":"f prime of x is this limit."},{"Start":"01:52.355 ","End":"01:54.860","Text":"Now, h is not 0 in this limit,"},{"Start":"01:54.860 ","End":"01:56.660","Text":"it\u0027s either positive or negative."},{"Start":"01:56.660 ","End":"02:02.330","Text":"If it\u0027s positive, then because f is strictly increasing,"},{"Start":"02:02.330 ","End":"02:08.390","Text":"then f of x plus h minus f of x is bigger than 0 because this is bigger than this."},{"Start":"02:08.390 ","End":"02:10.655","Text":"If h is negative,"},{"Start":"02:10.655 ","End":"02:16.785","Text":"then this minus this will be negative again from the increasing."},{"Start":"02:16.785 ","End":"02:22.850","Text":"In either case, this divided by h will be"},{"Start":"02:22.850 ","End":"02:28.640","Text":"positive because either have positive over positive or negative over negative."},{"Start":"02:28.640 ","End":"02:31.985","Text":"Now if this quotient is positive for each age,"},{"Start":"02:31.985 ","End":"02:33.710","Text":"when we take the limit,"},{"Start":"02:33.710 ","End":"02:36.500","Text":"it could be bigger or equal to 0."},{"Start":"02:36.500 ","End":"02:39.650","Text":"It could also be 0, but it can\u0027t be negative."},{"Start":"02:39.650 ","End":"02:42.550","Text":"That proves the 1 case."},{"Start":"02:42.550 ","End":"02:46.490","Text":"The other case, that if f is strictly decreasing,"},{"Start":"02:46.490 ","End":"02:49.655","Text":"then f prime less than or equal to 0."},{"Start":"02:49.655 ","End":"02:54.530","Text":"The second case is so similar that I\u0027ll just leave it,"},{"Start":"02:54.530 ","End":"02:55.850","Text":"it\u0027s practically the same,"},{"Start":"02:55.850 ","End":"03:01.650","Text":"just reversing some inequalities. We are done."}],"ID":26154},{"Watched":false,"Name":"Exercise 11","Duration":"2m 58s","ChapterTopicVideoID":25338,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25338.jpeg","UploadDate":"2021-04-01T16:18:48.6430000","DurationForVideoObject":"PT2M58S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.040","Text":"In this exercise, the function f is differentiable on the closed interval a,"},{"Start":"00:05.040 ","End":"00:12.870","Text":"b. Alpha is some number between f prime of a and f prime of b."},{"Start":"00:12.870 ","End":"00:16.755","Text":"We\u0027re assuming that f prime of a is less than f prime of b here."},{"Start":"00:16.755 ","End":"00:21.270","Text":"We define a second function, g, as follows."},{"Start":"00:21.270 ","End":"00:23.070","Text":"We have to show 2 things."},{"Start":"00:23.070 ","End":"00:28.305","Text":"First of all, that there is some c on the interval where g prime is 0,"},{"Start":"00:28.305 ","End":"00:30.240","Text":"and there\u0027s a hint."},{"Start":"00:30.240 ","End":"00:37.520","Text":"Part b is to use part a to conclude that if f is differentiable on an interval a,"},{"Start":"00:37.520 ","End":"00:43.655","Text":"b, then f prime has the intermediate value property on a, b."},{"Start":"00:43.655 ","End":"00:45.755","Text":"We\u0027ll start with part a,"},{"Start":"00:45.755 ","End":"00:47.225","Text":"which gives us a hint."},{"Start":"00:47.225 ","End":"00:52.670","Text":"G prime of a is f prime of a minus Alpha."},{"Start":"00:52.670 ","End":"00:55.970","Text":"Now, this just comes from differentiating this."},{"Start":"00:55.970 ","End":"01:02.090","Text":"G prime of b is f prime of b minus Alpha."},{"Start":"01:02.090 ","End":"01:05.810","Text":"This is negative since f prime of a is less than Alpha,"},{"Start":"01:05.810 ","End":"01:09.590","Text":"and this is positive since f prime of b is bigger than Alpha."},{"Start":"01:09.590 ","End":"01:14.150","Text":"Now we\u0027re going to use the hint to prove by contradiction."},{"Start":"01:14.150 ","End":"01:16.670","Text":"We suppose on the contrary,"},{"Start":"01:16.670 ","End":"01:21.725","Text":"that g prime is not equal to 0 anywhere on a, b,"},{"Start":"01:21.725 ","End":"01:28.490","Text":"which contradicts the existence of such c. Now by a previous problem,"},{"Start":"01:28.490 ","End":"01:32.250","Text":"and I\u0027ll show you what that previous problem is."},{"Start":"01:32.290 ","End":"01:37.080","Text":"Here it is, you can pause and read this."},{"Start":"01:37.130 ","End":"01:39.765","Text":"By this previous problem,"},{"Start":"01:39.765 ","End":"01:43.200","Text":"we get 1 of 2 possibilities."},{"Start":"01:43.200 ","End":"01:47.585","Text":"Either g prime is always bigger or equal to 0 on the interval,"},{"Start":"01:47.585 ","End":"01:51.445","Text":"or it\u0027s always less than or equal to 0 on the interval."},{"Start":"01:51.445 ","End":"01:58.985","Text":"This is a contradiction because we have that g prime is negative at a and positive at b,"},{"Start":"01:58.985 ","End":"02:02.615","Text":"so neither of these 2 possibilities can occur."},{"Start":"02:02.615 ","End":"02:07.280","Text":"By the way, this g prime of c is 0,"},{"Start":"02:07.280 ","End":"02:11.420","Text":"means that f prime of c is Alpha."},{"Start":"02:11.420 ","End":"02:13.835","Text":"We\u0027ll need this for part b."},{"Start":"02:13.835 ","End":"02:15.500","Text":"Onto b."},{"Start":"02:15.500 ","End":"02:18.710","Text":"Here we have 2 values, x_1 and x_2,"},{"Start":"02:18.710 ","End":"02:22.260","Text":"such that f of x_1 is less than Alpha,"},{"Start":"02:22.260 ","End":"02:23.895","Text":"less than f of x_2."},{"Start":"02:23.895 ","End":"02:27.570","Text":"I guess we\u0027re assuming that x_1 is less than x_2."},{"Start":"02:27.570 ","End":"02:29.520","Text":"Now by the above,"},{"Start":"02:29.520 ","End":"02:32.800","Text":"but with x_1, x_2 in place of a, b,"},{"Start":"02:32.800 ","End":"02:38.445","Text":"what we get is that there exists c in x_1,"},{"Start":"02:38.445 ","End":"02:40.840","Text":"x_2, which is a subset of a,"},{"Start":"02:40.840 ","End":"02:44.644","Text":"b such that g prime of c is 0,"},{"Start":"02:44.644 ","End":"02:49.165","Text":"meaning that f prime of c is Alpha."},{"Start":"02:49.165 ","End":"02:53.100","Text":"This actually demonstrates the intermediate value property,"},{"Start":"02:53.100 ","End":"02:58.180","Text":"if you think about it. We are done."}],"ID":26155},{"Watched":false,"Name":"Exercise 12","Duration":"2m 13s","ChapterTopicVideoID":25339,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25339.jpeg","UploadDate":"2021-04-01T16:19:12.4230000","DurationForVideoObject":"PT2M13S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.470","Text":"In this exercise, we have a differentiable function f on the closed interval a,b."},{"Start":"00:07.470 ","End":"00:11.130","Text":"We have to show that there are 3 points, c_1, c_2,"},{"Start":"00:11.130 ","End":"00:14.580","Text":"c_3 in the open interval a,b,"},{"Start":"00:14.580 ","End":"00:18.000","Text":"such that c_2 is not equal to c_3,"},{"Start":"00:18.000 ","End":"00:21.330","Text":"and this equation holds."},{"Start":"00:21.330 ","End":"00:24.800","Text":"Basically, what it says is that the derivative at c_1"},{"Start":"00:24.800 ","End":"00:28.670","Text":"is the average between the derivatives of c_2 and c_3."},{"Start":"00:28.670 ","End":"00:30.260","Text":"Let\u0027s look at the 3 points,"},{"Start":"00:30.260 ","End":"00:37.490","Text":"a,b and the midpoint a plus b over 2 and we\u0027ll apply the mean value theorem 3 times,"},{"Start":"00:37.490 ","End":"00:41.210","Text":"once with a,b once with these 2 and once with these 2."},{"Start":"00:41.210 ","End":"00:46.510","Text":"What we get is first 1 is that there is some c_1 between a and b,"},{"Start":"00:46.510 ","End":"00:50.700","Text":"such that this holds and the second is there\u0027s"},{"Start":"00:50.700 ","End":"00:56.535","Text":"some c_2 call it between a and a plus b over 2."},{"Start":"00:56.535 ","End":"01:00.360","Text":"Anyway, you could just read this what it says."},{"Start":"01:00.360 ","End":"01:02.980","Text":"It\u0027s just to, as I said, applying the mean value 3"},{"Start":"01:02.980 ","End":"01:06.115","Text":"times to each pair of points in these 3."},{"Start":"01:06.115 ","End":"01:10.805","Text":"Now if we add 2 plus 3,"},{"Start":"01:10.805 ","End":"01:12.684","Text":"I mean the equation numbers,"},{"Start":"01:12.684 ","End":"01:15.895","Text":"then the f of a plus b over 2 will cancel."},{"Start":"01:15.895 ","End":"01:18.580","Text":"We get f of b minus f of"},{"Start":"01:18.580 ","End":"01:26.145","Text":"a equals f prime of c_2 plus f prime of c_3 times this common part;"},{"Start":"01:26.145 ","End":"01:29.100","Text":"b minus a over 2, which I should have mentioned,"},{"Start":"01:29.100 ","End":"01:31.135","Text":"if you just do a bit of algebra,"},{"Start":"01:31.135 ","End":"01:32.845","Text":"this minus this is this,"},{"Start":"01:32.845 ","End":"01:35.795","Text":"and this minus this is the same thing."},{"Start":"01:35.795 ","End":"01:42.885","Text":"Then substituting f of b minus f of a from here on the left-hand side,"},{"Start":"01:42.885 ","End":"01:45.010","Text":"this is what we get."},{"Start":"01:45.010 ","End":"01:50.660","Text":"Notice that b minus a cancels."},{"Start":"01:50.660 ","End":"01:55.440","Text":"We have f prime of c_1 b minus a and"},{"Start":"01:55.440 ","End":"01:59.510","Text":"then here we pick the half in front and we also get a b minus a."},{"Start":"01:59.510 ","End":"02:02.915","Text":"What we\u0027re left with if we bring the 2 over to the left is that"},{"Start":"02:02.915 ","End":"02:07.870","Text":"twice f prime c_1 is this plus this."},{"Start":"02:07.870 ","End":"02:13.870","Text":"This is exactly what we had to show and so we are done."}],"ID":26156},{"Watched":false,"Name":"Exercise 13","Duration":"2m 36s","ChapterTopicVideoID":25340,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25340.jpeg","UploadDate":"2021-04-01T16:19:38.8030000","DurationForVideoObject":"PT2M36S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.400","Text":"In this exercise, we have a function from"},{"Start":"00:03.400 ","End":"00:07.955","Text":"the unit interval to the reals, and it\u0027s continuous."},{"Start":"00:07.955 ","End":"00:14.000","Text":"The integral from 0-1 of the function is 1."},{"Start":"00:14.000 ","End":"00:20.770","Text":"We have to show that there is some point c between 0 and 1 such that f of c equals 1."},{"Start":"00:20.770 ","End":"00:27.255","Text":"Then we have to show that there are 2 distinct c_1 and c_2 in this interval,"},{"Start":"00:27.255 ","End":"00:31.350","Text":"such that f of c_1 plus f of c_2 is 2."},{"Start":"00:31.350 ","End":"00:37.930","Text":"What we\u0027ll do is define a function F of x to be the integral from 0,"},{"Start":"00:37.930 ","End":"00:43.175","Text":"but just up to x. X will vary from 0-1."},{"Start":"00:43.175 ","End":"00:48.230","Text":"Yeah, as follows, F of x is integral from 0 to x of f of t dt."},{"Start":"00:48.230 ","End":"00:51.490","Text":"Now, note that when x is 0,"},{"Start":"00:51.490 ","End":"00:53.990","Text":"the integral 0-0 is 0,"},{"Start":"00:53.990 ","End":"00:57.760","Text":"and F of 1 is 1 follows from what is given here."},{"Start":"00:57.760 ","End":"01:01.840","Text":"Can apply the mean value theorem to F,"},{"Start":"01:01.840 ","End":"01:11.700","Text":"which is differentiable because the derivative of F is little f. There is a point c is 0,"},{"Start":"01:11.700 ","End":"01:17.070","Text":"1 such that F of 1 minus F of 0 is F-prime of c times 1 minus 0."},{"Start":"01:17.070 ","End":"01:20.335","Text":"Now, this is 1, like we said here,"},{"Start":"01:20.335 ","End":"01:22.685","Text":"and this is 0 like we said here."},{"Start":"01:22.685 ","End":"01:27.345","Text":"F-prime of c is little f of c,"},{"Start":"01:27.345 ","End":"01:32.300","Text":"and that is equal to 1. That\u0027s part a."},{"Start":"01:32.300 ","End":"01:36.560","Text":"In part b, we break it up into 2."},{"Start":"01:36.560 ","End":"01:40.880","Text":"I mean, we apply the mean value theorem separately on 0,"},{"Start":"01:40.880 ","End":"01:43.785","Text":"0.5 and then on 0.5, 1."},{"Start":"01:43.785 ","End":"01:50.040","Text":"What we\u0027ll get is that F of 0.5 minus F of 0 is f of c_1 0.5 minus 0,"},{"Start":"01:50.040 ","End":"01:54.350","Text":"and c_1 is in this interval and something similar in this interval,"},{"Start":"01:54.350 ","End":"01:59.285","Text":"notice that 1 minus 0.5 is a 0.5 and 0.5 minus 0 is also 0.5."},{"Start":"01:59.285 ","End":"02:04.350","Text":"What we get is that F of 1 minus F of 0,"},{"Start":"02:04.350 ","End":"02:06.540","Text":"we\u0027re adding these 2 equations."},{"Start":"02:06.540 ","End":"02:13.340","Text":"This bit F of 0.5 cancels equals this right-hand side plus this right-hand side."},{"Start":"02:13.340 ","End":"02:16.775","Text":"Like I said, this bit is 0.5 and this bit is 0.5."},{"Start":"02:16.775 ","End":"02:19.025","Text":"This is the expression we get."},{"Start":"02:19.025 ","End":"02:22.835","Text":"Now, if we multiply both sides by 2,"},{"Start":"02:22.835 ","End":"02:28.955","Text":"then we get that f of c_1 plus f of c_2 is twice,"},{"Start":"02:28.955 ","End":"02:30.885","Text":"and this is 1 minus 0, so it\u0027s 1."},{"Start":"02:30.885 ","End":"02:32.515","Text":"So we get this equals 2."},{"Start":"02:32.515 ","End":"02:34.475","Text":"That\u0027s part b."},{"Start":"02:34.475 ","End":"02:37.150","Text":"We are done."}],"ID":26157},{"Watched":false,"Name":"Exercise 14","Duration":"1m 41s","ChapterTopicVideoID":25341,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25341.jpeg","UploadDate":"2021-04-01T16:19:58.6970000","DurationForVideoObject":"PT1M41S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.860","Text":"In this exercise, we have a real valued function on the closed interval 0, 1."},{"Start":"00:04.860 ","End":"00:10.890","Text":"We\u0027re given that the absolute value of the derivative is less than 10"},{"Start":"00:10.890 ","End":"00:14.970","Text":"on the open interval excluding the end points."},{"Start":"00:14.970 ","End":"00:22.320","Text":"Now, we have a sequence x_n in the open interval which satisfies Cauchy\u0027s criterion."},{"Start":"00:22.320 ","End":"00:26.850","Text":"We have to show that f of x_n converges."},{"Start":"00:26.850 ","End":"00:30.720","Text":"Note we\u0027re not given that f is continuous or anything."},{"Start":"00:30.720 ","End":"00:37.780","Text":"The sequence satisfies the Cauchy criterion and this is what it is in technical terms."},{"Start":"00:37.780 ","End":"00:44.840","Text":"The distance between 2 elements goes to 0 when the indices get big enough."},{"Start":"00:44.840 ","End":"00:49.280","Text":"Fix n and m for the moment and for given n and m,"},{"Start":"00:49.280 ","End":"00:56.150","Text":"we can apply the mean value theorem and get this equation as here,"},{"Start":"00:56.150 ","End":"00:59.350","Text":"where c is between x_m and x_n."},{"Start":"00:59.350 ","End":"01:03.410","Text":"Taking the absolute value here, we get this"},{"Start":"01:03.410 ","End":"01:08.430","Text":"and now, we use the fact that this is less than 10,"},{"Start":"01:08.430 ","End":"01:11.480","Text":"so we get that this is less than or equal to this."},{"Start":"01:11.480 ","End":"01:14.420","Text":"This is true for each pair n and m,"},{"Start":"01:14.420 ","End":"01:18.920","Text":"and so when n, m goes to infinity,"},{"Start":"01:18.920 ","End":"01:22.440","Text":"the absolute value of x_n minus x_m goes to 0,"},{"Start":"01:22.440 ","End":"01:23.760","Text":"and if I multiply by 10,"},{"Start":"01:23.760 ","End":"01:25.200","Text":"it\u0027s still going to go to 0."},{"Start":"01:25.200 ","End":"01:26.790","Text":"10 times 0 is 0,"},{"Start":"01:26.790 ","End":"01:31.490","Text":"so this difference goes to 0 as n, m goes to infinity,"},{"Start":"01:31.490 ","End":"01:36.650","Text":"which means that f of x_n satisfies the Cauchy criterion too"},{"Start":"01:36.650 ","End":"01:38.960","Text":"and therefore it converges,"},{"Start":"01:38.960 ","End":"01:41.310","Text":"and we are done."}],"ID":26158},{"Watched":false,"Name":"Exercise 15","Duration":"2m 33s","ChapterTopicVideoID":25342,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25342.jpeg","UploadDate":"2021-04-01T16:20:27.0430000","DurationForVideoObject":"PT2M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.875","Text":"In this exercise, we have a function from the closed unit interval to itself."},{"Start":"00:05.875 ","End":"00:11.575","Text":"It has the property that the derivative is negative in all the interval."},{"Start":"00:11.575 ","End":"00:17.000","Text":"We have to show that there is exactly 1 c in this intervals,"},{"Start":"00:17.000 ","End":"00:20.740","Text":"which satisfies f of c equals c squared."},{"Start":"00:20.740 ","End":"00:23.650","Text":"Let\u0027s look at the difference between these 2."},{"Start":"00:23.650 ","End":"00:28.345","Text":"We\u0027ll call it big F of x and it\u0027s f of x minus x squared."},{"Start":"00:28.345 ","End":"00:32.245","Text":"Then, big F of 0 is bigger or equal to 0."},{"Start":"00:32.245 ","End":"00:38.710","Text":"Because x squared, when x is 0 is 0 and f of x is in 0,"},{"Start":"00:38.710 ","End":"00:40.855","Text":"1, so it\u0027s bigger or equal to 0."},{"Start":"00:40.855 ","End":"00:42.890","Text":"In the case of 1,"},{"Start":"00:42.890 ","End":"00:46.105","Text":"we have something between 0 and 1 and we subtract 1,"},{"Start":"00:46.105 ","End":"00:48.410","Text":"so it has to be less than or equal to 0."},{"Start":"00:48.410 ","End":"00:53.490","Text":"Now, using the IVP, Intermediate Value Property,"},{"Start":"00:53.490 ","End":"00:58.715","Text":"because of this bigger or equal to and less than or equal to, somewhere in between,"},{"Start":"00:58.715 ","End":"01:00.545","Text":"we have a point,"},{"Start":"01:00.545 ","End":"01:05.000","Text":"call it x_1, where f of x_1 is exactly equal to 0."},{"Start":"01:05.000 ","End":"01:08.510","Text":"That means if you interpret it here,"},{"Start":"01:08.510 ","End":"01:11.940","Text":"that f of x_1 equals x_1 squared,"},{"Start":"01:11.940 ","End":"01:14.520","Text":"so we found 1 solution."},{"Start":"01:14.520 ","End":"01:16.275","Text":"We take c equals x_1."},{"Start":"01:16.275 ","End":"01:18.620","Text":"I\u0027m not sure there can\u0027t be 2 solutions,"},{"Start":"01:18.620 ","End":"01:20.675","Text":"so do that by contradiction."},{"Start":"01:20.675 ","End":"01:22.415","Text":"Suppose that there\u0027s another 1,"},{"Start":"01:22.415 ","End":"01:25.190","Text":"x_2 different from the first 1,"},{"Start":"01:25.190 ","End":"01:27.440","Text":"and also satisfies the equation,"},{"Start":"01:27.440 ","End":"01:30.940","Text":"so f of x_2 equals x_2 squared,"},{"Start":"01:30.940 ","End":"01:34.430","Text":"and we can apply the mean value theorem to x_1 and"},{"Start":"01:34.430 ","End":"01:38.630","Text":"x_2 and conclude that there\u0027s some c between them,"},{"Start":"01:38.630 ","End":"01:45.615","Text":"such that f of x_1 minus f of x_2 is f prime of c x_1 minus x_2,"},{"Start":"01:45.615 ","End":"01:47.400","Text":"using the mean value theorem."},{"Start":"01:47.400 ","End":"01:53.730","Text":"Now, f of x_1 is equal to x_1 squared and f of x_2 equals x_2 squared,"},{"Start":"01:53.730 ","End":"01:54.915","Text":"so we get this."},{"Start":"01:54.915 ","End":"01:58.600","Text":"Now, divide both sides by x_1 minus x_2,"},{"Start":"01:58.600 ","End":"02:01.580","Text":"and using the difference of squares formula,"},{"Start":"02:01.580 ","End":"02:04.714","Text":"this comes out to be just x_1 plus x_2,"},{"Start":"02:04.714 ","End":"02:08.000","Text":"and this is positive because x_1"},{"Start":"02:08.000 ","End":"02:11.540","Text":"and x_2 are in the interval from 0-1 and they\u0027re different."},{"Start":"02:11.540 ","End":"02:15.784","Text":"This is a contradiction, because in fact,"},{"Start":"02:15.784 ","End":"02:21.610","Text":"f prime is strictly negative for all x in this interval."},{"Start":"02:21.610 ","End":"02:26.090","Text":"This contradiction means that we can\u0027t have another 1,"},{"Start":"02:26.090 ","End":"02:33.630","Text":"so there is just 1 solution to this equation. We\u0027re done."}],"ID":26159},{"Watched":false,"Name":"Exercise 16","Duration":"3m 43s","ChapterTopicVideoID":25343,"CourseChapterTopicPlaylistID":257166,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25343.jpeg","UploadDate":"2021-04-01T16:21:14.4430000","DurationForVideoObject":"PT3M43S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.500","Text":"In this exercise, f is a function defined on"},{"Start":"00:04.500 ","End":"00:11.520","Text":"the closed unit interval and use it to define a sequence a_n as follows,"},{"Start":"00:11.520 ","End":"00:14.940","Text":"a_n equals f of 1 over n minus f of 1 over n plus"},{"Start":"00:14.940 ","End":"00:18.890","Text":"1 for each n. We have to show first of all that,"},{"Start":"00:18.890 ","End":"00:20.780","Text":"if f is continuous,"},{"Start":"00:20.780 ","End":"00:25.635","Text":"then the series with a_n converges,"},{"Start":"00:25.635 ","End":"00:29.600","Text":"and the second thing is that if f is differentiable"},{"Start":"00:29.600 ","End":"00:34.970","Text":"and if the derivative in absolute value is less than a 1/2,"},{"Start":"00:34.970 ","End":"00:40.070","Text":"then this series converges."},{"Start":"00:40.070 ","End":"00:45.020","Text":"In part 1, we have the sum of an infinite series,"},{"Start":"00:45.020 ","End":"00:50.270","Text":"and that is defined to be the limit of the partial sums."},{"Start":"00:50.270 ","End":"00:55.850","Text":"In other words, if we let S_n be the sum starting from 1 to infinity,"},{"Start":"00:55.850 ","End":"00:59.600","Text":"from 1 to n, then we have to show that that converges."},{"Start":"00:59.600 ","End":"01:03.424","Text":"Now, this is a telescoping series,"},{"Start":"01:03.424 ","End":"01:06.020","Text":"f of 1 minus f of a 1/2 plus f of"},{"Start":"01:06.020 ","End":"01:09.875","Text":"a 1/2 minus f of a 1/3 plus f of a 1/3 minus f of a 1/4."},{"Start":"01:09.875 ","End":"01:13.520","Text":"You see all the middle terms cancel and we\u0027re just left with"},{"Start":"01:13.520 ","End":"01:17.805","Text":"f of 1 minus f of 1 over n plus 1,"},{"Start":"01:17.805 ","End":"01:25.205","Text":"and the limit of the partial sum is therefore f of 1 minus f of 0."},{"Start":"01:25.205 ","End":"01:29.660","Text":"This is from the continuity of f because what\u0027s inside goes to 0,"},{"Start":"01:29.660 ","End":"01:32.255","Text":"f of it goes to f of 0."},{"Start":"01:32.255 ","End":"01:37.610","Text":"This infinite series converges to f of 1 minus f of 0."},{"Start":"01:37.610 ","End":"01:39.760","Text":"Now in part 2, f is differentiable,"},{"Start":"01:39.760 ","End":"01:48.620","Text":"so we can apply the mean value theorem to the interval from 1 over n plus 1 to 1 over n,"},{"Start":"01:48.620 ","End":"01:55.610","Text":"and we get that there exists some point in this interval such that f of 1 over n minus"},{"Start":"01:55.610 ","End":"02:03.155","Text":"f of 1 over n plus 1 is f prime of c times 1 over n minus 1 over n plus 1,"},{"Start":"02:03.155 ","End":"02:05.660","Text":"and this is a_n."},{"Start":"02:05.660 ","End":"02:08.630","Text":"Applying the absolute value to both sides,"},{"Start":"02:08.630 ","End":"02:16.895","Text":"we get that the absolute value of a_n is equal to f prime of c_n in absolute value,"},{"Start":"02:16.895 ","End":"02:18.770","Text":"which is less than a half,"},{"Start":"02:18.770 ","End":"02:20.059","Text":"and this is positive,"},{"Start":"02:20.059 ","End":"02:23.300","Text":"so this just stays as is."},{"Start":"02:23.300 ","End":"02:25.460","Text":"This is equal to,"},{"Start":"02:25.460 ","End":"02:29.405","Text":"if you just figure out this fraction subtraction."},{"Start":"02:29.405 ","End":"02:31.360","Text":"Get n plus 1 minus n,"},{"Start":"02:31.360 ","End":"02:34.150","Text":"which is 1 over n times n plus 1,"},{"Start":"02:34.150 ","End":"02:36.065","Text":"and put the 2 there."},{"Start":"02:36.065 ","End":"02:39.275","Text":"Now if we take the absolute value of this expression,"},{"Start":"02:39.275 ","End":"02:42.995","Text":"this will be less than what we had for a_n,"},{"Start":"02:42.995 ","End":"02:48.289","Text":"and then the absolute value of Cosine n is less than or equal to 1,"},{"Start":"02:48.289 ","End":"02:51.245","Text":"and the square root of n goes here."},{"Start":"02:51.245 ","End":"02:53.765","Text":"We get this is less than this,"},{"Start":"02:53.765 ","End":"02:58.280","Text":"and now we can estimate this by replacing n plus 1 with n,"},{"Start":"02:58.280 ","End":"03:02.640","Text":"decrease the denominator, so you increase the fraction,"},{"Start":"03:02.640 ","End":"03:06.435","Text":"this is less than the 1/2 was here,"},{"Start":"03:06.435 ","End":"03:09.995","Text":"and then here we have root n over n squared."},{"Start":"03:09.995 ","End":"03:13.880","Text":"It\u0027s n to the power of minus 3 over 2,"},{"Start":"03:13.880 ","End":"03:16.520","Text":"and this is a p series."},{"Start":"03:16.520 ","End":"03:20.000","Text":"This is like 1 over n to the power of 3 over 2."},{"Start":"03:20.000 ","End":"03:23.920","Text":"It\u0027s a p series with p equals 3 over 2,"},{"Start":"03:23.920 ","End":"03:26.670","Text":"and whenever the p is bigger than 1,"},{"Start":"03:26.670 ","End":"03:28.140","Text":"then the series converges."},{"Start":"03:28.140 ","End":"03:30.810","Text":"The 1/2 doesn\u0027t affect the convergence."},{"Start":"03:30.810 ","End":"03:39.220","Text":"If this converges, then so does the sum of this because there\u0027s a comparison test."},{"Start":"03:39.220 ","End":"03:44.020","Text":"We\u0027ve proved that this converges, so we\u0027re done."}],"ID":26160}],"Thumbnail":null,"ID":257166},{"Name":"Darboux\u0027s Mean Value Theorem","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Darboux\u0027s Mean Value Theorem","Duration":"5m 44s","ChapterTopicVideoID":25360,"CourseChapterTopicPlaylistID":257167,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25360.jpeg","UploadDate":"2021-04-01T16:52:12.2070000","DurationForVideoObject":"PT5M44S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.175","Text":"In this clip, we\u0027ll learn another mean value theorem due to Darboux,"},{"Start":"00:05.175 ","End":"00:08.175","Text":"usually just called Darboux\u0027s theorem."},{"Start":"00:08.175 ","End":"00:11.790","Text":"Not to be confused with Lagrange\u0027s mean value theorem,"},{"Start":"00:11.790 ","End":"00:14.595","Text":"which is the regular mean value theorem."},{"Start":"00:14.595 ","End":"00:16.380","Text":"What does the theorem say?"},{"Start":"00:16.380 ","End":"00:20.340","Text":"Let f be differentiable on the closed interval a, b"},{"Start":"00:20.340 ","End":"00:26.310","Text":"and let Gamma be between f prime of a and f prime of b."},{"Start":"00:26.310 ","End":"00:31.980","Text":"F prime exists because f is differentiable on the whole interval."},{"Start":"00:31.980 ","End":"00:34.184","Text":"When I say between,"},{"Start":"00:34.184 ","End":"00:38.700","Text":"this is what it is in terms of inequalities."},{"Start":"00:38.700 ","End":"00:41.310","Text":"Its 2 ways it could be, is f prime of a,"},{"Start":"00:41.310 ","End":"00:45.240","Text":"it could be less than or equal to f prime of b or vice versa."},{"Start":"00:45.250 ","End":"00:48.064","Text":"If this is the case,"},{"Start":"00:48.064 ","End":"00:52.790","Text":"then there exists c in the interval a, b"},{"Start":"00:52.790 ","End":"00:58.020","Text":"such that f prime of c is exactly equal to Gamma."},{"Start":"00:58.240 ","End":"01:05.315","Text":"The asterisks here just means that we should maybe strictly say that"},{"Start":"01:05.315 ","End":"01:09.830","Text":"the derivative at a is the right derivative"},{"Start":"01:09.830 ","End":"01:14.080","Text":"and the derivative at b is the left derivative."},{"Start":"01:14.080 ","End":"01:16.920","Text":"We won\u0027t prove this theorem."},{"Start":"01:16.920 ","End":"01:20.035","Text":"As they say, it\u0027s beyond the scope of this course."},{"Start":"01:20.035 ","End":"01:23.060","Text":"Now the theorem implies that the function,"},{"Start":"01:23.060 ","End":"01:25.220","Text":"which is derivative on an interval,"},{"Start":"01:25.220 ","End":"01:29.900","Text":"has the intermediate value property on the interval."},{"Start":"01:29.900 ","End":"01:32.555","Text":"That means that it maps intervals to intervals."},{"Start":"01:32.555 ","End":"01:34.535","Text":"There\u0027s no gaps."},{"Start":"01:34.535 ","End":"01:38.165","Text":"If you look at what it says like here, I mean,"},{"Start":"01:38.165 ","End":"01:41.795","Text":"if you have 2 numbers in the image,"},{"Start":"01:41.795 ","End":"01:43.985","Text":"f prime of a and f prime of b,"},{"Start":"01:43.985 ","End":"01:47.780","Text":"then any number between them is also in the image"},{"Start":"01:47.780 ","End":"01:52.145","Text":"and the Gamma is realized as f prime of c,"},{"Start":"01:52.145 ","End":"01:54.845","Text":"so f prime has no gaps."},{"Start":"01:54.845 ","End":"01:58.490","Text":"Now of course, if the derivative were continuous,"},{"Start":"01:58.490 ","End":"02:02.570","Text":"the result would have followed from the intermediate value theorem"},{"Start":"02:02.570 ","End":"02:06.680","Text":"because any continuous function has the intermediate value property."},{"Start":"02:06.680 ","End":"02:11.990","Text":"Darboux proves it also when we don\u0027t know that the derivative is continuous,"},{"Start":"02:11.990 ","End":"02:13.535","Text":"just that it exists."},{"Start":"02:13.535 ","End":"02:19.760","Text":"Second remark is that the fact that f is differentiable on an interval"},{"Start":"02:19.760 ","End":"02:26.540","Text":"restricts the discontinuities that its derivative can have quite severely."},{"Start":"02:26.540 ","End":"02:29.000","Text":"That\u0027s what Darboux\u0027s theorem does."},{"Start":"02:29.000 ","End":"02:35.015","Text":"For example, it cannot have a removable discontinuity."},{"Start":"02:35.015 ","End":"02:38.510","Text":"It can\u0027t have a type 1 discontinuity,"},{"Start":"02:38.510 ","End":"02:41.975","Text":"which is a jump discontinuity where the left and right limits."},{"Start":"02:41.975 ","End":"02:44.510","Text":"They both exist, but they\u0027re not equal."},{"Start":"02:44.510 ","End":"02:49.520","Text":"It can\u0027t have the subtype 2 discontinuity,"},{"Start":"02:49.520 ","End":"02:54.940","Text":"where the left or right limit is infinity."},{"Start":"02:54.940 ","End":"02:58.820","Text":"None of the following can be a derivative."},{"Start":"02:58.820 ","End":"03:01.250","Text":"It\u0027s going to be example of each one of these."},{"Start":"03:01.250 ","End":"03:05.210","Text":"A of x, which is 4x or x minus 1,"},{"Start":"03:05.210 ","End":"03:09.605","Text":"has a jump discontinuity at x equals 1."},{"Start":"03:09.605 ","End":"03:15.995","Text":"This has a removable discontinuity at x equals 1."},{"Start":"03:15.995 ","End":"03:18.080","Text":"I mean, x equals 1 its 0."},{"Start":"03:18.080 ","End":"03:19.880","Text":"But near x equals 1,"},{"Start":"03:19.880 ","End":"03:22.375","Text":"the function is near 4."},{"Start":"03:22.375 ","End":"03:25.035","Text":"The other 1, C of x,"},{"Start":"03:25.035 ","End":"03:28.485","Text":"that has an infinite discontinuity."},{"Start":"03:28.485 ","End":"03:33.530","Text":"It doesn\u0027t matter what we define it at 0 itself, but when it\u0027s near 0,"},{"Start":"03:33.530 ","End":"03:37.610","Text":"the values get close to plus or minus infinity,"},{"Start":"03:37.610 ","End":"03:41.315","Text":"to actually infinity on the right and minus infinity on the left."},{"Start":"03:41.315 ","End":"03:46.100","Text":"We\u0027ll actually prove these 3 in the exercises."},{"Start":"03:46.100 ","End":"03:52.715","Text":"Now what remains if you remove all these 3 types of discontinuity?"},{"Start":"03:52.715 ","End":"04:02.615","Text":"The only one that\u0027s left is a subtype of 2 where the left and right limit don\u0027t exist."},{"Start":"04:02.615 ","End":"04:10.800","Text":"Now an example is take f of x to be x squared sine 1 over x, except at 0."},{"Start":"04:10.800 ","End":"04:12.140","Text":"We wouldn\u0027t be define to there."},{"Start":"04:12.140 ","End":"04:14.570","Text":"We let it be whatever we want, say 0."},{"Start":"04:14.570 ","End":"04:17.210","Text":"Turns out this is differentiable for all x."},{"Start":"04:17.210 ","End":"04:20.930","Text":"It\u0027s a famous example that\u0027s used in mathematics"},{"Start":"04:20.930 ","End":"04:27.020","Text":"and its derivative turns out to be discontinuous at 0."},{"Start":"04:27.020 ","End":"04:34.640","Text":"But the discontinuity it has is that the limit on the right doesn\u0027t exist."},{"Start":"04:34.640 ","End":"04:36.200","Text":"Well, also on the left."},{"Start":"04:36.200 ","End":"04:37.070","Text":"Let\u0027s see."},{"Start":"04:37.070 ","End":"04:38.240","Text":"The limit on the right,"},{"Start":"04:38.240 ","End":"04:40.340","Text":"when x goes to 0,"},{"Start":"04:40.340 ","End":"04:43.355","Text":"this part goes to 0."},{"Start":"04:43.355 ","End":"04:47.825","Text":"That\u0027s okay because it\u0027s something goes to 0 times something bounded."},{"Start":"04:47.825 ","End":"04:50.960","Text":"But this part, 1 over x,"},{"Start":"04:50.960 ","End":"04:56.165","Text":"goes to infinity and the cosine keeps oscillating plus or minus 1."},{"Start":"04:56.165 ","End":"05:02.405","Text":"It oscillates as we get to 0 and never approaches any particular limit."},{"Start":"05:02.405 ","End":"05:04.640","Text":"Now, the converse is not true."},{"Start":"05:04.640 ","End":"05:13.055","Text":"What I mean is here we showed that we had a derivative with a subtype 2 discontinuity,"},{"Start":"05:13.055 ","End":"05:16.985","Text":"where the limit on the right or could left doesn\u0027t exist."},{"Start":"05:16.985 ","End":"05:21.235","Text":"But there could be a function with such a discontinuity,"},{"Start":"05:21.235 ","End":"05:23.240","Text":"which isn\u0027t a derivative."},{"Start":"05:23.240 ","End":"05:29.690","Text":"In other words, if you just know that a function that has this type of discontinuity,"},{"Start":"05:29.690 ","End":"05:33.070","Text":"you can\u0027t tell whether it is or isn\u0027t a derivative."},{"Start":"05:33.070 ","End":"05:38.465","Text":"An example of the other kind is what is called the Dirichlet function,"},{"Start":"05:38.465 ","End":"05:42.445","Text":"but that\u0027s dealt within one of the exercises."},{"Start":"05:42.445 ","End":"05:45.480","Text":"That\u0027s all for this introduction."}],"ID":26176},{"Watched":false,"Name":"Exercise 1","Duration":"1m 53s","ChapterTopicVideoID":25361,"CourseChapterTopicPlaylistID":257167,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25361.jpeg","UploadDate":"2021-04-01T16:52:38.3430000","DurationForVideoObject":"PT1M53S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.139","Text":"In this exercise, there\u0027s a question,"},{"Start":"00:04.139 ","End":"00:08.220","Text":"does there exist a differentiable function f such that"},{"Start":"00:08.220 ","End":"00:14.190","Text":"the derivative of f is equal to 4x if x is less than 1"},{"Start":"00:14.190 ","End":"00:17.235","Text":"and x minus 1 when x is bigger or equal to 1."},{"Start":"00:17.235 ","End":"00:19.320","Text":"Let\u0027s draw a picture of that."},{"Start":"00:19.320 ","End":"00:21.360","Text":"Up to the point 1,"},{"Start":"00:21.360 ","End":"00:23.040","Text":"we have the line 4x,"},{"Start":"00:23.040 ","End":"00:25.500","Text":"and from x equals 1,"},{"Start":"00:25.500 ","End":"00:28.215","Text":"the line x minus 1."},{"Start":"00:28.215 ","End":"00:30.180","Text":"There\u0027s a jump here,"},{"Start":"00:30.180 ","End":"00:34.530","Text":"and we mentioned in the tutorial that a derivative can\u0027t have a jump discontinuity."},{"Start":"00:34.530 ","End":"00:38.610","Text":"Let\u0027s prove this directly from the Darboux\u0027s theorem."},{"Start":"00:38.610 ","End":"00:41.335","Text":"Let\u0027s suppose proof by contradiction,"},{"Start":"00:41.335 ","End":"00:44.290","Text":"that such a function f exists."},{"Start":"00:44.290 ","End":"00:49.905","Text":"We\u0027ll use Darboux\u0027s\u0027 theorem on the interval from 0.5 to 1."},{"Start":"00:49.905 ","End":"00:52.675","Text":"Notice that at 1, it\u0027s equal to 0,"},{"Start":"00:52.675 ","End":"00:55.595","Text":"and at 0.5 it\u0027s equal to 2."},{"Start":"00:55.595 ","End":"00:59.855","Text":"The plan is to take a point in between 2 and 0, which is 1,"},{"Start":"00:59.855 ","End":"01:04.220","Text":"and see if any x in this interval that can give us an output of 1."},{"Start":"01:04.220 ","End":"01:07.535","Text":"You see for the picture there isn\u0027t so like I said,"},{"Start":"01:07.535 ","End":"01:14.020","Text":"f prime of 0.5 is 2 and f prime of 1 is 0 just from here or from the picture."},{"Start":"01:14.020 ","End":"01:16.800","Text":"Now we choose a number between 0 and 2 and like I said,"},{"Start":"01:16.800 ","End":"01:18.720","Text":"1 is the obvious choice."},{"Start":"01:18.720 ","End":"01:24.835","Text":"By the theorem, there has to be some c in this interval such that f prime of c is 1."},{"Start":"01:24.835 ","End":"01:27.045","Text":"You can see from the picture doesn\u0027t work there."},{"Start":"01:27.045 ","End":"01:29.400","Text":"Here\u0027s 1 and here\u0027s the function."},{"Start":"01:29.400 ","End":"01:32.675","Text":"They don\u0027t agree or we can say, suppose there is,"},{"Start":"01:32.675 ","End":"01:37.475","Text":"in this portion, we\u0027re going by this formula and the 4x part."},{"Start":"01:37.475 ","End":"01:39.380","Text":"We have 4c equals 1."},{"Start":"01:39.380 ","End":"01:41.315","Text":"So c is a 1/4,"},{"Start":"01:41.315 ","End":"01:45.170","Text":"and a 1/4 is not in the interval from 0.5 to 1"},{"Start":"01:45.170 ","End":"01:47.740","Text":"and that\u0027s a contradiction."},{"Start":"01:47.740 ","End":"01:50.610","Text":"Contradiction came from assuming that f exists,"},{"Start":"01:50.610 ","End":"01:53.920","Text":"so f doesn\u0027t exist and we\u0027re done."}],"ID":26177},{"Watched":false,"Name":"Exercise 2","Duration":"1m 37s","ChapterTopicVideoID":25353,"CourseChapterTopicPlaylistID":257167,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25353.jpeg","UploadDate":"2021-04-01T16:42:01.1470000","DurationForVideoObject":"PT1M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.449","Text":"In this exercise, you are asked to decide"},{"Start":"00:03.449 ","End":"00:07.305","Text":"whether or not there exists a differentiable function f,"},{"Start":"00:07.305 ","End":"00:11.085","Text":"such that the derivative of f is the following."},{"Start":"00:11.085 ","End":"00:17.325","Text":"It\u0027s equal to 4 when x is 0 and x squared when x is not 0."},{"Start":"00:17.325 ","End":"00:19.930","Text":"Here\u0027s the picture of what it looks like."},{"Start":"00:19.930 ","End":"00:25.350","Text":"Notice there\u0027s a missing point on the curve here and it\u0027s placed over here."},{"Start":"00:25.350 ","End":"00:28.620","Text":"This is a removable discontinuity."},{"Start":"00:28.620 ","End":"00:32.660","Text":"The answer is no. You might remember in the tutorial we said that a"},{"Start":"00:32.660 ","End":"00:36.500","Text":"derivative can\u0027t have a removable discontinuity,"},{"Start":"00:36.500 ","End":"00:39.470","Text":"but we\u0027ll prove it directly from Darboux\u0027s theorem."},{"Start":"00:39.470 ","End":"00:40.880","Text":"I\u0027ll prove it by contradiction,"},{"Start":"00:40.880 ","End":"00:42.430","Text":"by assuming it does exist,"},{"Start":"00:42.430 ","End":"00:44.260","Text":"and see what happens."},{"Start":"00:44.260 ","End":"00:49.445","Text":"By Darboux\u0027s theorem, and we\u0027ll choose the interval from 0-1,"},{"Start":"00:49.445 ","End":"00:51.395","Text":"what we have is at 0,"},{"Start":"00:51.395 ","End":"00:53.795","Text":"the value is 4, at 1,"},{"Start":"00:53.795 ","End":"00:55.810","Text":"the value is 1."},{"Start":"00:55.810 ","End":"01:00.740","Text":"We\u0027re going to choose something between 4 and 1, let\u0027s say 2."},{"Start":"01:00.740 ","End":"01:05.665","Text":"Now, Darboux\u0027s theorem would say that somewhere between 0 and 1,"},{"Start":"01:05.665 ","End":"01:09.620","Text":"the curve should get the value 2 but we see it doesn\u0027t."},{"Start":"01:09.620 ","End":"01:11.435","Text":"Let\u0027s do this more formally."},{"Start":"01:11.435 ","End":"01:12.650","Text":"By the theorem, there is a c,"},{"Start":"01:12.650 ","End":"01:14.435","Text":"this f prime of c is 2,"},{"Start":"01:14.435 ","End":"01:16.940","Text":"so c-squared equals 2 because we\u0027re in"},{"Start":"01:16.940 ","End":"01:20.240","Text":"the portion where we\u0027re using the formula x squared,"},{"Start":"01:20.240 ","End":"01:25.370","Text":"we\u0027re not at 0. So c is root 2 or minus root 2."},{"Start":"01:25.370 ","End":"01:31.585","Text":"But in any case, it\u0027s not in the interval from 0 to 1 and that\u0027s a contradiction."},{"Start":"01:31.585 ","End":"01:33.920","Text":"The contradiction came from assuming that f exists"},{"Start":"01:33.920 ","End":"01:37.230","Text":"and therefore it doesn\u0027t, and we\u0027re done."}],"ID":26169},{"Watched":false,"Name":"Exercise 3","Duration":"1m 43s","ChapterTopicVideoID":25354,"CourseChapterTopicPlaylistID":257167,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25354.jpeg","UploadDate":"2021-04-01T16:42:20.1230000","DurationForVideoObject":"PT1M43S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.900","Text":"In this exercise, we\u0027re asked to decide whether or not there exists"},{"Start":"00:03.900 ","End":"00:10.004","Text":"a differentiable function f whose derivative is given by the following formula."},{"Start":"00:10.004 ","End":"00:12.590","Text":"At x equals 0, it\u0027s equal to 1,"},{"Start":"00:12.590 ","End":"00:14.595","Text":"and when x is not 0,"},{"Start":"00:14.595 ","End":"00:18.075","Text":"we can define it as 1 over x squared."},{"Start":"00:18.075 ","End":"00:22.770","Text":"Let\u0027s have a picture to see what\u0027s happening here."},{"Start":"00:22.770 ","End":"00:25.620","Text":"We see it\u0027s an infinite discontinuity."},{"Start":"00:25.620 ","End":"00:28.410","Text":"0, we\u0027ve defined it artificially as 1,"},{"Start":"00:28.410 ","End":"00:30.300","Text":"but on either side, it goes to infinity,"},{"Start":"00:30.300 ","End":"00:34.920","Text":"and you might remember that derivatives can\u0027t have infinite discontinuities,"},{"Start":"00:34.920 ","End":"00:38.130","Text":"but we\u0027re going to prove this directly from Darboux\u0027s theorem,"},{"Start":"00:38.130 ","End":"00:40.215","Text":"and we\u0027ll do it by contradiction,"},{"Start":"00:40.215 ","End":"00:43.275","Text":"supposing that such an f exists."},{"Start":"00:43.275 ","End":"00:46.249","Text":"Darboux\u0027s theorem will help us reach a contradiction."},{"Start":"00:46.249 ","End":"00:49.985","Text":"Look at it on the interval from 0 to 0.5."},{"Start":"00:49.985 ","End":"00:51.470","Text":"Now, at the endpoint, at 0,"},{"Start":"00:51.470 ","End":"00:54.140","Text":"it\u0027s equal to 1; at 0.5,"},{"Start":"00:54.140 ","End":"00:56.485","Text":"it\u0027s equal to 4."},{"Start":"00:56.485 ","End":"01:01.185","Text":"Now, we\u0027ll choose a number between 1 and 4, let\u0027s say 2,"},{"Start":"01:01.185 ","End":"01:04.430","Text":"and we can see what the problem is because we\u0027re supposed"},{"Start":"01:04.430 ","End":"01:08.720","Text":"to take on any value between 1 and 4 in this interval,"},{"Start":"01:08.720 ","End":"01:14.885","Text":"but there is nowhere in this interval that can be 2 because it\u0027s all way up here."},{"Start":"01:14.885 ","End":"01:18.240","Text":"By the theorem, there exists a c in this interval."},{"Start":"01:18.240 ","End":"01:19.965","Text":"F prime of c is 2."},{"Start":"01:19.965 ","End":"01:22.690","Text":"This is the graph of f-prime, of course."},{"Start":"01:23.210 ","End":"01:25.410","Text":"If there was a c in here,"},{"Start":"01:25.410 ","End":"01:27.060","Text":"we\u0027d have 1 over c squared,"},{"Start":"01:27.060 ","End":"01:28.560","Text":"from the formula here, equals 2,"},{"Start":"01:28.560 ","End":"01:30.465","Text":"so c is 1 over root 2,"},{"Start":"01:30.465 ","End":"01:36.200","Text":"and 1 over root 2 is not in the interval from 0 to a 1/2,"},{"Start":"01:36.200 ","End":"01:43.590","Text":"so we\u0027ve gotten a contradiction which means that f doesn\u0027t exist, and we\u0027re done."}],"ID":26170},{"Watched":false,"Name":"Exercise 4","Duration":"5m 22s","ChapterTopicVideoID":25355,"CourseChapterTopicPlaylistID":257167,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25355.jpeg","UploadDate":"2021-04-01T16:43:42.0970000","DurationForVideoObject":"PT5M22S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.895","Text":"This exercise has 2 parts."},{"Start":"00:02.895 ","End":"00:05.130","Text":"The first part is theoretical."},{"Start":"00:05.130 ","End":"00:09.885","Text":"We\u0027re given a function from an interval to the reals,"},{"Start":"00:09.885 ","End":"00:14.085","Text":"and we\u0027re told it\u0027s differentiable and we take a point inside the interval."},{"Start":"00:14.085 ","End":"00:20.445","Text":"We have to show that that point can\u0027t be a removable discontinuity of f prime,"},{"Start":"00:20.445 ","End":"00:22.740","Text":"we mentioned this in the tutorial."},{"Start":"00:22.740 ","End":"00:25.350","Text":"Part b is based on part a to show that"},{"Start":"00:25.350 ","End":"00:31.275","Text":"the following function cannot be the derivative of another function."},{"Start":"00:31.275 ","End":"00:34.515","Text":"In other words, we can\u0027t have f such that f prime is the following."},{"Start":"00:34.515 ","End":"00:36.135","Text":"Start with the picture,"},{"Start":"00:36.135 ","End":"00:38.835","Text":"and we\u0027re going to do this by contradiction,"},{"Start":"00:38.835 ","End":"00:41.530","Text":"so we\u0027ll assume that x naught,"},{"Start":"00:41.530 ","End":"00:43.160","Text":"which is between a and b,"},{"Start":"00:43.160 ","End":"00:47.045","Text":"is a removable discontinuity and we\u0027ll reach a contradiction at the end."},{"Start":"00:47.045 ","End":"00:50.450","Text":"By the definition of removable discontinuity,"},{"Start":"00:50.450 ","End":"00:56.750","Text":"there exists the limit as x goes to x naught f prime of x,"},{"Start":"00:56.750 ","End":"01:02.705","Text":"and its sum L, but that is not equal to the value of f prime at the point."},{"Start":"01:02.705 ","End":"01:05.960","Text":"We\u0027re going to assume that L is less than."},{"Start":"01:05.960 ","End":"01:08.840","Text":"We thought equal is less than or bigger than,"},{"Start":"01:08.840 ","End":"01:13.130","Text":"and the proof is almost the same in both cases."},{"Start":"01:13.130 ","End":"01:16.520","Text":"You could also use a trick to generalize by just"},{"Start":"01:16.520 ","End":"01:20.755","Text":"considering the function minus f instead of f,"},{"Start":"01:20.755 ","End":"01:24.405","Text":"and it will still be differentiable,"},{"Start":"01:24.405 ","End":"01:28.550","Text":"and still have a removable discontinuity at the same place,"},{"Start":"01:28.550 ","End":"01:31.640","Text":"but the inequality would be reversed."},{"Start":"01:31.640 ","End":"01:37.365","Text":"Anyway, the definition of limit says that for all Epsilon,"},{"Start":"01:37.365 ","End":"01:40.350","Text":"there is a Delta such that,"},{"Start":"01:40.350 ","End":"01:44.580","Text":"if x is close to x naught within Delta and it\u0027s not equal to 0,"},{"Start":"01:44.580 ","End":"01:46.905","Text":"it can\u0027t be the point x naught itself,"},{"Start":"01:46.905 ","End":"01:53.045","Text":"then the distance from f prime to L is less than Epsilon."},{"Start":"01:53.045 ","End":"01:56.470","Text":"That\u0027s the regular Epsilon Delta definition."},{"Start":"01:56.470 ","End":"01:59.625","Text":"The next thing is to choose our Epsilon,"},{"Start":"01:59.625 ","End":"02:08.335","Text":"and we\u0027ll choose Epsilon to be the gap between here and here, divided by 2."},{"Start":"02:08.335 ","End":"02:12.860","Text":"Because we took the case where L is less than f prime,"},{"Start":"02:12.860 ","End":"02:14.285","Text":"this will be positive."},{"Start":"02:14.285 ","End":"02:17.270","Text":"If we add this Epsilon to L,"},{"Start":"02:17.270 ","End":"02:19.400","Text":"we only get halfway here."},{"Start":"02:19.400 ","End":"02:22.075","Text":"In any event, it\u0027s less than f prime of x naught,"},{"Start":"02:22.075 ","End":"02:27.620","Text":"and here\u0027s a more formal way of writing this out, but it\u0027s obvious."},{"Start":"02:27.620 ","End":"02:30.560","Text":"Let Delta correspond to the Epsilon,"},{"Start":"02:30.560 ","End":"02:34.025","Text":"as in this Epsilon Delta guarantee."},{"Start":"02:34.025 ","End":"02:37.265","Text":"We choose t in this interval,"},{"Start":"02:37.265 ","End":"02:44.215","Text":"and then because t minus x naught absolute value is less than Delta,"},{"Start":"02:44.215 ","End":"02:50.210","Text":"then we get by this definition that f prime of t minus L is less than Epsilon,"},{"Start":"02:50.210 ","End":"02:53.345","Text":"and so f prime of t is less than L plus Epsilon,"},{"Start":"02:53.345 ","End":"02:55.340","Text":"just by interpreting this,"},{"Start":"02:55.340 ","End":"02:57.860","Text":"which is less than f prime of x naught."},{"Start":"02:57.860 ","End":"03:01.910","Text":"What we\u0027re going to do is to apply Darboux\u0027s theorem to"},{"Start":"03:01.910 ","End":"03:06.500","Text":"the interval from x naught to this t that we chose."},{"Start":"03:06.500 ","End":"03:12.450","Text":"We can choose a Gamma such that Gamma is between L plus Epsilon and f prime of x naught."},{"Start":"03:12.450 ","End":"03:13.550","Text":"This is less than this,"},{"Start":"03:13.550 ","End":"03:16.840","Text":"there\u0027s plenty of numbers in between, just choose 1."},{"Start":"03:16.840 ","End":"03:20.820","Text":"Then, f prime of t is less than Gamma,"},{"Start":"03:20.820 ","End":"03:26.130","Text":"is less than f prime of x naught because f prime of t is less than L plus Epsilon,"},{"Start":"03:26.130 ","End":"03:28.840","Text":"replacing this by something less."},{"Start":"03:29.360 ","End":"03:34.850","Text":"The theorem then guarantees that we have some point c in"},{"Start":"03:34.850 ","End":"03:40.410","Text":"this open interval such that f prime of c is exactly this Gamma."},{"Start":"03:40.970 ","End":"03:44.060","Text":"Because c is in this interval,"},{"Start":"03:44.060 ","End":"03:49.930","Text":"the distance from c to x naught is less than the distance from t to x naught,"},{"Start":"03:49.930 ","End":"03:54.690","Text":"so we get 0 is less than this, less than Delta,"},{"Start":"03:54.690 ","End":"03:57.175","Text":"and from this Delta Epsilon,"},{"Start":"03:57.175 ","End":"04:01.330","Text":"we get that f prime of c minus L is less than Epsilon."},{"Start":"04:01.330 ","End":"04:04.640","Text":"Gamma, which is f prime of c,"},{"Start":"04:04.640 ","End":"04:06.425","Text":"is less than L plus Epsilon."},{"Start":"04:06.425 ","End":"04:08.935","Text":"Just bring this to the other side."},{"Start":"04:08.935 ","End":"04:12.710","Text":"We have that Gamma\u0027s less than L plus Epsilon,"},{"Start":"04:12.710 ","End":"04:15.470","Text":"and I highlighted this in red,"},{"Start":"04:15.470 ","End":"04:16.880","Text":"because on the other hand,"},{"Start":"04:16.880 ","End":"04:20.635","Text":"we chose Gamma bigger than L plus Epsilon."},{"Start":"04:20.635 ","End":"04:22.400","Text":"Here, Gamma is bigger than,"},{"Start":"04:22.400 ","End":"04:25.080","Text":"so it can\u0027t be both bigger than and less than,"},{"Start":"04:25.080 ","End":"04:27.290","Text":"and this is the current prediction we\u0027re looking for,"},{"Start":"04:27.290 ","End":"04:34.765","Text":"and that settles part a. F prime of a function can\u0027t have a removable discontinuity."},{"Start":"04:34.765 ","End":"04:39.140","Text":"In part b, we\u0027re given a function whose derivative is this,"},{"Start":"04:39.140 ","End":"04:41.480","Text":"we had this in a previous exercise."},{"Start":"04:41.480 ","End":"04:45.740","Text":"This has a removable discontinuity at x equals naught,"},{"Start":"04:45.740 ","End":"04:51.935","Text":"because the limit as x goes to 0 of f prime of x,"},{"Start":"04:51.935 ","End":"04:55.010","Text":"is equal to limit of x squared because x"},{"Start":"04:55.010 ","End":"04:58.190","Text":"is not equal to 0 when it approaches 0, and not 0."},{"Start":"04:58.190 ","End":"05:01.850","Text":"On the other hand, f prime of 0 is 4."},{"Start":"05:01.850 ","End":"05:05.270","Text":"This limit is not equal to the value at the point,"},{"Start":"05:05.270 ","End":"05:07.850","Text":"and so it has a removable discontinuity,"},{"Start":"05:07.850 ","End":"05:09.860","Text":"but we just showed that it can\u0027t have"},{"Start":"05:09.860 ","End":"05:14.095","Text":"a removable discontinuity if it\u0027s f prime of something."},{"Start":"05:14.095 ","End":"05:16.370","Text":"Which means that f doesn\u0027t exist,"},{"Start":"05:16.370 ","End":"05:20.380","Text":"because its mere existence gives us a contradiction."},{"Start":"05:20.380 ","End":"05:23.450","Text":"That\u0027s the end of this exercise."}],"ID":26171},{"Watched":false,"Name":"Exercise 5","Duration":"8m ","ChapterTopicVideoID":25356,"CourseChapterTopicPlaylistID":257167,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25356.jpeg","UploadDate":"2021-04-01T16:46:16.0370000","DurationForVideoObject":"PT8M","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.370","Text":"In this exercise in part a,"},{"Start":"00:02.370 ","End":"00:06.840","Text":"we\u0027re going to prove the results that we already stated that the derivative of"},{"Start":"00:06.840 ","End":"00:12.195","Text":"a function can\u0027t have a type 1 or jump discontinuity,"},{"Start":"00:12.195 ","End":"00:15.330","Text":"and in part b, we\u0027ll use this to show"},{"Start":"00:15.330 ","End":"00:19.260","Text":"that the function given here can\u0027t be the derivative,"},{"Start":"00:19.260 ","End":"00:23.010","Text":"otherwise there is just no f such that f-prime is this function here."},{"Start":"00:23.010 ","End":"00:28.455","Text":"I can already tell you that this is going to be an example with a jump discontinuity."},{"Start":"00:28.455 ","End":"00:31.260","Text":"Let\u0027s start with part a, the proof."},{"Start":"00:31.260 ","End":"00:35.135","Text":"We\u0027ll do it by contradiction by supposing that x-Naught,"},{"Start":"00:35.135 ","End":"00:36.755","Text":"which is between a and b,"},{"Start":"00:36.755 ","End":"00:40.810","Text":"is a type 1 discontinuity of f-prime,"},{"Start":"00:40.810 ","End":"00:44.370","Text":"or f-prime has the type 1 discontinuity at x-Naught."},{"Start":"00:44.370 ","End":"00:45.560","Text":"Like in the picture,"},{"Start":"00:45.560 ","End":"00:47.090","Text":"the limit on the left exists,"},{"Start":"00:47.090 ","End":"00:48.680","Text":"and the limit on the right exists,"},{"Start":"00:48.680 ","End":"00:50.045","Text":"but they\u0027re not the same,"},{"Start":"00:50.045 ","End":"00:52.865","Text":"regardless of what happens at the point itself."},{"Start":"00:52.865 ","End":"01:00.195","Text":"The left limit of the derivative is just the left derivative,"},{"Start":"01:00.195 ","End":"01:04.085","Text":"and the right limit of the derivative is just the right derivative."},{"Start":"01:04.085 ","End":"01:06.590","Text":"What we\u0027re saying is that these 2 both exist,"},{"Start":"01:06.590 ","End":"01:07.880","Text":"but are not equal."},{"Start":"01:07.880 ","End":"01:09.755","Text":"If these 2 are not equal,"},{"Start":"01:09.755 ","End":"01:11.945","Text":"then 1 is bigger than the other,"},{"Start":"01:11.945 ","End":"01:14.010","Text":"and it\u0027s entirely symmetrical,"},{"Start":"01:14.010 ","End":"01:19.520","Text":"so we can say that the left derivative is less than the right derivative,"},{"Start":"01:19.520 ","End":"01:21.560","Text":"and we write without loss of generality,"},{"Start":"01:21.560 ","End":"01:24.880","Text":"meaning it\u0027s just the same argument to do it the other way."},{"Start":"01:24.880 ","End":"01:28.355","Text":"If you don\u0027t like that, there\u0027s another trick we can do."},{"Start":"01:28.355 ","End":"01:31.685","Text":"We can just say that if we proved it for this case,"},{"Start":"01:31.685 ","End":"01:35.195","Text":"now consider g, which is f of minus x,"},{"Start":"01:35.195 ","End":"01:40.215","Text":"and that will also have a type 1 discontinuity,"},{"Start":"01:40.215 ","End":"01:44.705","Text":"and it will satisfy the inequality the other way around."},{"Start":"01:44.705 ","End":"01:48.980","Text":"Let Epsilon be 1/3 of the gap"},{"Start":"01:48.980 ","End":"01:53.165","Text":"between this right limit and the left limit of the derivative,"},{"Start":"01:53.165 ","End":"02:00.140","Text":"so that this left plus Epsilon is strictly less than the right minus Epsilon."},{"Start":"02:00.140 ","End":"02:03.010","Text":"In fact, the difference is bigger than Epsilon."},{"Start":"02:03.010 ","End":"02:07.190","Text":"Let\u0027s go back to the equivalent expression."},{"Start":"02:07.190 ","End":"02:08.630","Text":"Instead of the right derivative,"},{"Start":"02:08.630 ","End":"02:12.980","Text":"we can replace it with the right limit of the derivative."},{"Start":"02:12.980 ","End":"02:18.865","Text":"Similarly, the left derivative is the limit to the left of the derivative."},{"Start":"02:18.865 ","End":"02:23.510","Text":"What we get using Epsilon Delta on the left,"},{"Start":"02:23.510 ","End":"02:32.580","Text":"we get a Delta_1 such that if x is between x-Naught minus Delta and x-Naught,"},{"Start":"02:32.580 ","End":"02:35.155","Text":"it\u0027s just a bit to the left here,"},{"Start":"02:35.155 ","End":"02:41.225","Text":"then this absolute value of the difference is less than Epsilon."},{"Start":"02:41.225 ","End":"02:43.295","Text":"Similarly on the right,"},{"Start":"02:43.295 ","End":"02:44.930","Text":"we have a Delta-2,"},{"Start":"02:44.930 ","End":"02:46.460","Text":"with a similar condition,"},{"Start":"02:46.460 ","End":"02:49.610","Text":"I\u0027ll you pause and look at the small details."},{"Start":"02:49.610 ","End":"02:53.890","Text":"What we\u0027ll do is we\u0027ll take Delta to be the smaller of the 2,"},{"Start":"02:53.890 ","End":"02:58.005","Text":"then if you look at the Delta neighborhood of x-Naught,"},{"Start":"02:58.005 ","End":"03:02.220","Text":"this is contained in x-Naught minus Delta_1, x-Naught plus Delta_2,"},{"Start":"03:02.220 ","End":"03:05.490","Text":"because Delta is less than or equal to Delta_1,"},{"Start":"03:05.490 ","End":"03:07.340","Text":"also less than or equal to Delta_2,"},{"Start":"03:07.340 ","End":"03:10.490","Text":"and this is contained in our interval a,"},{"Start":"03:10.490 ","End":"03:15.590","Text":"b, so that this is inside of this, everything is defined,"},{"Start":"03:15.590 ","End":"03:19.220","Text":"and we can choose Alpha and Beta such"},{"Start":"03:19.220 ","End":"03:23.420","Text":"that Alpha is sandwiched between x-Naught minus Delta,"},{"Start":"03:23.420 ","End":"03:29.410","Text":"and x-Naught, and Beta is sandwiched between x-Naught and x-Naught plus Delta."},{"Start":"03:29.410 ","End":"03:34.640","Text":"By this here, by the left limit,"},{"Start":"03:34.640 ","End":"03:37.145","Text":"replacing x by Alpha here,"},{"Start":"03:37.145 ","End":"03:39.695","Text":"we get this, and similarly,"},{"Start":"03:39.695 ","End":"03:45.349","Text":"if we look at this, because Beta satisfies this condition,"},{"Start":"03:45.349 ","End":"03:48.170","Text":"so we can replace this x here by Beta,"},{"Start":"03:48.170 ","End":"03:51.970","Text":"we\u0027ve got this, and from these,"},{"Start":"03:51.970 ","End":"03:57.065","Text":"just by throwing away the absolute value and choosing which way around to subtract,"},{"Start":"03:57.065 ","End":"04:01.390","Text":"we\u0027ve got that this is less than this plus Epsilon,"},{"Start":"04:01.390 ","End":"04:03.920","Text":"and here we can reverse"},{"Start":"04:03.920 ","End":"04:06.950","Text":"the order of the subtraction before throwing out the absolute value,"},{"Start":"04:06.950 ","End":"04:09.920","Text":"and then rearrange and we get this."},{"Start":"04:09.920 ","End":"04:13.565","Text":"Look at this, we\u0027re going to use that."},{"Start":"04:13.565 ","End":"04:17.915","Text":"What we had above is what\u0027s written here, then from here,"},{"Start":"04:17.915 ","End":"04:20.960","Text":"we can put f-prime of Alpha here, and because of this,"},{"Start":"04:20.960 ","End":"04:23.455","Text":"we can put an f-prime of Beta here,"},{"Start":"04:23.455 ","End":"04:26.015","Text":"and here I just copied it."},{"Start":"04:26.015 ","End":"04:28.400","Text":"Because of this less than,"},{"Start":"04:28.400 ","End":"04:32.960","Text":"in-between these 2, we can choose a number, call it Gamma,"},{"Start":"04:32.960 ","End":"04:35.705","Text":"which is between this and this,"},{"Start":"04:35.705 ","End":"04:39.005","Text":"but because this is bigger than this and this is less,"},{"Start":"04:39.005 ","End":"04:42.050","Text":"this Gamma is also between these 2."},{"Start":"04:42.050 ","End":"04:48.230","Text":"I forgot to say, we have to choose it so that Gamma is not equal to f-prime of x-Naught."},{"Start":"04:48.230 ","End":"04:50.720","Text":"In the interval between this and this,"},{"Start":"04:50.720 ","End":"04:53.030","Text":"there\u0027s plenty of numbers, infinitely many,"},{"Start":"04:53.030 ","End":"04:58.440","Text":"so we can choose 1 and make sure it\u0027s not equal to f-prime of x-Naught."},{"Start":"04:58.440 ","End":"05:00.950","Text":"As I said, yeah, Gammas is also between here and here,"},{"Start":"05:00.950 ","End":"05:02.510","Text":"because that\u0027s a wider interval,"},{"Start":"05:02.510 ","End":"05:09.640","Text":"and we applied our boost theorem to f on the interval from a-b."},{"Start":"05:09.640 ","End":"05:13.850","Text":"We have that there exists some c between Alpha and Beta"},{"Start":"05:13.850 ","End":"05:18.215","Text":"such that f-prime of c is exactly equal to Gamma."},{"Start":"05:18.215 ","End":"05:22.789","Text":"Notice that c is not equal to x-Naught,"},{"Start":"05:22.789 ","End":"05:26.480","Text":"because then we would have that f-prime of x-Naught equals Gamma,"},{"Start":"05:26.480 ","End":"05:29.620","Text":"and we chose it so as not to be equal."},{"Start":"05:29.620 ","End":"05:32.090","Text":"Now, a not equal means there are 2 cases."},{"Start":"05:32.090 ","End":"05:34.460","Text":"It could be bigger than or smaller than."},{"Start":"05:34.460 ","End":"05:36.875","Text":"Let\u0027s take each case separately."},{"Start":"05:36.875 ","End":"05:38.930","Text":"We reach a contradiction in each case."},{"Start":"05:38.930 ","End":"05:42.205","Text":"If c is less than x-Naught,"},{"Start":"05:42.205 ","End":"05:44.699","Text":"then c is bigger than Alpha,"},{"Start":"05:44.699 ","End":"05:47.880","Text":"and Alpha is bigger than x-Naught minus Delta,"},{"Start":"05:47.880 ","End":"05:51.060","Text":"and that\u0027s bigger or equal to x-Naught minus Delta_1,"},{"Start":"05:51.060 ","End":"05:53.215","Text":"because Delta is less than or equal Delta_1."},{"Start":"05:53.215 ","End":"06:00.150","Text":"That means that c is in the interval from x-Naught minus Delta_1 to x-Naught."},{"Start":"06:00.150 ","End":"06:01.995","Text":"It\u0027s less than x-Naught,"},{"Start":"06:01.995 ","End":"06:03.600","Text":"and from this chain,"},{"Start":"06:03.600 ","End":"06:04.925","Text":"if you just take the ends,"},{"Start":"06:04.925 ","End":"06:06.965","Text":"c is bigger than this,"},{"Start":"06:06.965 ","End":"06:08.425","Text":"yeah, it\u0027s in this interval,"},{"Start":"06:08.425 ","End":"06:14.360","Text":"by condition 1 above regarding the limit on the left,"},{"Start":"06:14.360 ","End":"06:17.510","Text":"we get that this is less than Epsilon,"},{"Start":"06:17.510 ","End":"06:24.425","Text":"and that means that Gamma minus left derivative at x-Naught is less than Epsilon."},{"Start":"06:24.425 ","End":"06:27.410","Text":"On the other hand, you have the less than,"},{"Start":"06:27.410 ","End":"06:29.120","Text":"we\u0027re going to also show that it\u0027s bigger than,"},{"Start":"06:29.120 ","End":"06:32.165","Text":"and that\u0027s because if we look here,"},{"Start":"06:32.165 ","End":"06:35.830","Text":"we have this plus Epsilon is less than Gamma,"},{"Start":"06:35.830 ","End":"06:45.725","Text":"so we can just rearrange and say that Epsilon is less than Gamma minus this,"},{"Start":"06:45.725 ","End":"06:47.960","Text":"and just turn the sides around."},{"Start":"06:47.960 ","End":"06:50.410","Text":"We have a contradiction."},{"Start":"06:50.410 ","End":"06:53.915","Text":"I\u0027ll let you look at the other mirror image,"},{"Start":"06:53.915 ","End":"06:56.315","Text":"at the other right-hand case,"},{"Start":"06:56.315 ","End":"06:58.190","Text":"and we get something very similar."},{"Start":"06:58.190 ","End":"07:03.435","Text":"We get a contradiction this way also, almost identically here."},{"Start":"07:03.435 ","End":"07:07.320","Text":"That means that either way we get a contradiction,"},{"Start":"07:07.320 ","End":"07:13.150","Text":"and that means that the assumption that such an f exists is false."},{"Start":"07:13.150 ","End":"07:19.115","Text":"F can\u0027t exist where its derivative has a jump discontinuity."},{"Start":"07:19.115 ","End":"07:21.470","Text":"Part b is just an application of this."},{"Start":"07:21.470 ","End":"07:25.070","Text":"We already did this longhand in the previous exercise, but here,"},{"Start":"07:25.070 ","End":"07:28.910","Text":"we can do it immediately by noticing that"},{"Start":"07:28.910 ","End":"07:35.105","Text":"this function has a jump discontinuity at x equals 1."},{"Start":"07:35.105 ","End":"07:38.090","Text":"The limit on the right,"},{"Start":"07:38.090 ","End":"07:40.325","Text":"we take it from here,"},{"Start":"07:40.325 ","End":"07:42.110","Text":"and when x goes to 1,"},{"Start":"07:42.110 ","End":"07:43.475","Text":"it comes out 2."},{"Start":"07:43.475 ","End":"07:49.185","Text":"But the limit on the left is 4 times 1, which is 4,"},{"Start":"07:49.185 ","End":"07:50.790","Text":"and 2 is not equal to 4,"},{"Start":"07:50.790 ","End":"07:53.330","Text":"so left limit and right limit are not equal,"},{"Start":"07:53.330 ","End":"07:56.255","Text":"so it\u0027s a jump discontinuity."},{"Start":"07:56.255 ","End":"08:01.560","Text":"It\u0027s a contradiction to what we just showed in part a. Now we\u0027re done."}],"ID":26172},{"Watched":false,"Name":"Exercise 6","Duration":"4m 46s","ChapterTopicVideoID":25357,"CourseChapterTopicPlaylistID":257167,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25357.jpeg","UploadDate":"2021-04-01T16:48:23.1870000","DurationForVideoObject":"PT4M46S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"In this exercise, basically,"},{"Start":"00:02.280 ","End":"00:05.520","Text":"what we\u0027re asked to do is to show that"},{"Start":"00:05.520 ","End":"00:09.959","Text":"if a function is a derivative of another function on an interval,"},{"Start":"00:09.959 ","End":"00:13.515","Text":"then it can\u0027t have an infinite discontinuity there."},{"Start":"00:13.515 ","End":"00:15.165","Text":"You can read it in full."},{"Start":"00:15.165 ","End":"00:20.580","Text":"In part b, we\u0027re asked to decide whether a specific example is differentiable or not."},{"Start":"00:20.580 ","End":"00:24.270","Text":"It turns out we can answer immediately using part a,"},{"Start":"00:24.270 ","End":"00:28.260","Text":"though we have had this in previous exercise."},{"Start":"00:28.260 ","End":"00:32.700","Text":"Let\u0027s suppose, and we\u0027ll do this by contradiction,"},{"Start":"00:32.700 ","End":"00:37.020","Text":"that x-naught is an infinite discontinuity"},{"Start":"00:37.020 ","End":"00:39.235","Text":"where x-naught is between a and b,"},{"Start":"00:39.235 ","End":"00:43.545","Text":"and let\u0027s say that it\u0027s the right derivative."},{"Start":"00:43.545 ","End":"00:45.710","Text":"We\u0027re going to reach a contradiction."},{"Start":"00:45.710 ","End":"00:52.175","Text":"Now, there are 3 other cases because we have left and right to plus or minus infinity."},{"Start":"00:52.175 ","End":"00:57.500","Text":"But we can just say without loss of generality or similarly for the other cases,"},{"Start":"00:57.500 ","End":"00:58.850","Text":"or if you don\u0027t like that,"},{"Start":"00:58.850 ","End":"01:02.450","Text":"you can consider the function f of minus x"},{"Start":"01:02.450 ","End":"01:05.810","Text":"and that will switch left and right."},{"Start":"01:05.810 ","End":"01:09.215","Text":"Consider also plus or minus f of x,"},{"Start":"01:09.215 ","End":"01:13.100","Text":"in either case, to get the infinity to minus infinity."},{"Start":"01:13.100 ","End":"01:15.410","Text":"In other words, we can either say similarly"},{"Start":"01:15.410 ","End":"01:19.625","Text":"or we can just adjust the functions with minuses inside or outside."},{"Start":"01:19.625 ","End":"01:22.325","Text":"We\u0027re taking the infinite case on the right."},{"Start":"01:22.325 ","End":"01:23.870","Text":"This is what we\u0027re claiming."},{"Start":"01:23.870 ","End":"01:27.290","Text":"That\u0027s the same as saying that, actually needs proving,"},{"Start":"01:27.290 ","End":"01:31.400","Text":"but it is equal to the limit as x goes to x-naught"},{"Start":"01:31.400 ","End":"01:34.730","Text":"from the right of the 2 side of derivative."},{"Start":"01:34.730 ","End":"01:39.725","Text":"We\u0027ll use the Delta M definition of limit at infinity."},{"Start":"01:39.725 ","End":"01:45.440","Text":"Suppose M is large and we need it to be bigger than f prime of x-naught."},{"Start":"01:45.440 ","End":"01:46.550","Text":"It\u0027s going to go to infinity,"},{"Start":"01:46.550 ","End":"01:49.735","Text":"so at some point it\u0027s going to be bigger than f prime of x-naught."},{"Start":"01:49.735 ","End":"01:54.035","Text":"Then there exists a corresponding Delta bigger than 0."},{"Start":"01:54.035 ","End":"01:56.060","Text":"But again, we need an extra restriction."},{"Start":"01:56.060 ","End":"02:01.640","Text":"We can always make Delta smaller than b minus x-naught if we need such that,"},{"Start":"02:01.640 ","End":"02:05.375","Text":"if x is in this interval,"},{"Start":"02:05.375 ","End":"02:07.190","Text":"we don\u0027t need the 2-sided interval,"},{"Start":"02:07.190 ","End":"02:11.420","Text":"we just need the right side interval from x-naught to x-naught plus Delta,"},{"Start":"02:11.420 ","End":"02:14.300","Text":"which happens to be inside a, b."},{"Start":"02:14.300 ","End":"02:16.375","Text":"That\u0027s because of this condition."},{"Start":"02:16.375 ","End":"02:20.660","Text":"If this holds, then f prime is bigger than M,"},{"Start":"02:20.660 ","End":"02:25.850","Text":"and now choose a particular one of these x\u0027s in this interval and call it x_1."},{"Start":"02:25.850 ","End":"02:27.685","Text":"Just fix 1 of these."},{"Start":"02:27.685 ","End":"02:33.405","Text":"We\u0027re going to use Darboux\u0027s theorem on the interval from x-naught to x_1."},{"Start":"02:33.405 ","End":"02:38.570","Text":"Note, x_1 is 1 of these x\u0027s where f prime of x is bigger than M."},{"Start":"02:38.570 ","End":"02:41.095","Text":"So f prime of x_1 is bigger than M,"},{"Start":"02:41.095 ","End":"02:42.465","Text":"and also from here,"},{"Start":"02:42.465 ","End":"02:44.190","Text":"f prime of x-naught is less than M,"},{"Start":"02:44.190 ","End":"02:46.920","Text":"so we have that M is between these 2."},{"Start":"02:46.920 ","End":"02:50.330","Text":"Now we can choose a Gamma between this and this."},{"Start":"02:50.330 ","End":"02:52.940","Text":"Gamma is between f prime of x-naught and M."},{"Start":"02:52.940 ","End":"02:58.815","Text":"That makes Gamma also between f prime of x-naught and f prime of x_1."},{"Start":"02:58.815 ","End":"03:04.430","Text":"By the theorem, c exists between x-naught and x_1"},{"Start":"03:04.430 ","End":"03:08.200","Text":"such that f prime of c is equal to Gamma."},{"Start":"03:08.200 ","End":"03:11.540","Text":"Now, the interval x-naught, x_1 is contained"},{"Start":"03:11.540 ","End":"03:14.150","Text":"in the interval x-naught to x-naught plus Delta"},{"Start":"03:14.150 ","End":"03:18.525","Text":"because x_1 is less than x-naught plus Delta."},{"Start":"03:18.525 ","End":"03:21.565","Text":"Here it is, x_1 is in this interval."},{"Start":"03:21.565 ","End":"03:26.100","Text":"This implies that f prime of c is bigger than M."},{"Start":"03:26.100 ","End":"03:27.095","Text":"What\u0027s the reason?"},{"Start":"03:27.095 ","End":"03:30.005","Text":"We saw that whenever x is in this interval,"},{"Start":"03:30.005 ","End":"03:32.210","Text":"f prime of x is bigger than M."},{"Start":"03:32.210 ","End":"03:35.630","Text":"In particular, you could take x equals c."},{"Start":"03:35.630 ","End":"03:39.605","Text":"Well, c is in this interval, so c is in this interval,"},{"Start":"03:39.605 ","End":"03:42.635","Text":"and so f prime of c is bigger than M,"},{"Start":"03:42.635 ","End":"03:46.050","Text":"so Gamma is bigger than M."},{"Start":"03:46.050 ","End":"03:49.310","Text":"Now, here we have a contradiction because on the one hand,"},{"Start":"03:49.310 ","End":"03:52.550","Text":"Gamma, which is f prime of c is bigger than M."},{"Start":"03:52.550 ","End":"03:57.660","Text":"On the other hand, we already showed that Gamma is less than M."},{"Start":"03:57.660 ","End":"03:58.500","Text":"Here it is."},{"Start":"03:58.500 ","End":"04:02.790","Text":"Gamma is less than M, so it can\u0027t be bigger than M and less than M."},{"Start":"04:02.790 ","End":"04:04.310","Text":"That\u0027s a contradiction."},{"Start":"04:04.310 ","End":"04:08.750","Text":"This contradiction came from assuming the existence of such a function f."},{"Start":"04:08.750 ","End":"04:12.710","Text":"We can\u0027t have any infinite discontinuities,"},{"Start":"04:12.710 ","End":"04:16.230","Text":"plus or minus, left or right, we can\u0027t have 1."},{"Start":"04:16.230 ","End":"04:19.250","Text":"In part b, it\u0027s an example we had before"},{"Start":"04:19.250 ","End":"04:22.255","Text":"where we take the function 1 over x squared,"},{"Start":"04:22.255 ","End":"04:25.190","Text":"except at 0, we define it however you like."},{"Start":"04:25.190 ","End":"04:27.830","Text":"It has an infinite discontinuity on either side."},{"Start":"04:27.830 ","End":"04:30.935","Text":"When x goes to 0, either on the left or on the right,"},{"Start":"04:30.935 ","End":"04:35.135","Text":"then f prime of x here goes to infinity."},{"Start":"04:35.135 ","End":"04:38.284","Text":"In particular, the case that we proved,"},{"Start":"04:38.284 ","End":"04:40.670","Text":"and that\u0027s a contradiction."},{"Start":"04:40.670 ","End":"04:46.830","Text":"The answer is no, such a function f does not exist whose derivative is this."}],"ID":26173},{"Watched":false,"Name":"Exercise 7","Duration":"3m 33s","ChapterTopicVideoID":25358,"CourseChapterTopicPlaylistID":257167,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25358.jpeg","UploadDate":"2021-04-01T16:49:42.1230000","DurationForVideoObject":"PT3M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.440","Text":"In this exercise, f is continuous on"},{"Start":"00:04.440 ","End":"00:09.840","Text":"the closed unit interval and it\u0027s differentiable on the open unit interval."},{"Start":"00:09.840 ","End":"00:15.810","Text":"We\u0027re given that the limit from the right of f prime of"},{"Start":"00:15.810 ","End":"00:21.840","Text":"x is L. That\u0027s as x goes to 0 from the right,"},{"Start":"00:21.840 ","End":"00:24.270","Text":"where L is some real number."},{"Start":"00:24.270 ","End":"00:28.655","Text":"We have to show that the right derivative at 0"},{"Start":"00:28.655 ","End":"00:33.200","Text":"is also L. Part b is essentially the same as Part a,"},{"Start":"00:33.200 ","End":"00:37.475","Text":"except that we replace L by plus or minus infinity."},{"Start":"00:37.475 ","End":"00:43.385","Text":"Let\u0027s start with Part a and we\u0027ll work from the definition of Right derivative."},{"Start":"00:43.385 ","End":"00:49.280","Text":"It\u0027s like the regular derivative except that the limit is not h goes to 0,"},{"Start":"00:49.280 ","End":"00:52.880","Text":"but h goes to 0 plus meaning 0 from the right."},{"Start":"00:52.880 ","End":"00:55.580","Text":"We have to show this exists, this limit."},{"Start":"00:55.580 ","End":"01:00.559","Text":"So it\u0027s f of h minus f naught over h. In general,"},{"Start":"01:00.559 ","End":"01:06.530","Text":"if instead of 0 we had a would be f of a plus h minus f of a over h, the limit."},{"Start":"01:06.530 ","End":"01:08.885","Text":"We\u0027re going to use L\u0027Hospital\u0027s rule,"},{"Start":"01:08.885 ","End":"01:16.100","Text":"that f of x be little f of x minus f of 0 and G of x equals x."},{"Start":"01:16.100 ","End":"01:19.100","Text":"The idea is that when h goes to 0 here,"},{"Start":"01:19.100 ","End":"01:22.550","Text":"the numerator goes to 0 and the denominator goes to 0."},{"Start":"01:22.550 ","End":"01:25.365","Text":"So that motivates the L\u0027Hospital\u0027s."},{"Start":"01:25.365 ","End":"01:26.630","Text":"If we define it this way,"},{"Start":"01:26.630 ","End":"01:30.560","Text":"what we have in the numerator is big F of h,"},{"Start":"01:30.560 ","End":"01:38.370","Text":"but we have in the denominator is big G of h. We want to show that this limit exists and"},{"Start":"01:38.370 ","End":"01:42.110","Text":"what L\u0027Hospital\u0027s says is that if we"},{"Start":"01:42.110 ","End":"01:46.310","Text":"have a 0 over 0 situation and the numerator goes to 0,"},{"Start":"01:46.310 ","End":"01:47.960","Text":"denominator goes to 0,"},{"Start":"01:47.960 ","End":"01:52.280","Text":"then it\u0027s sufficient to show that this limit"},{"Start":"01:52.280 ","End":"01:56.420","Text":"of the derivative of F over the derivative of G,"},{"Start":"01:56.420 ","End":"01:58.805","Text":"if this exists and is equal to L,"},{"Start":"01:58.805 ","End":"02:02.450","Text":"then the original limit also is equal to"},{"Start":"02:02.450 ","End":"02:07.070","Text":"L. That\u0027s the L\u0027Hospital\u0027s rule for the 0 over 0 case."},{"Start":"02:07.070 ","End":"02:14.060","Text":"Now, F prime of H is equal to f prime of h minus 0."},{"Start":"02:14.060 ","End":"02:16.175","Text":"And if we differentiate this,"},{"Start":"02:16.175 ","End":"02:18.560","Text":"this little f prime and this is a constant,"},{"Start":"02:18.560 ","End":"02:25.280","Text":"so that goes to 0, and the denominator G of x is x so G prime is just 1, the constant."},{"Start":"02:25.280 ","End":"02:34.730","Text":"So this quotient becomes just F prime of H. We need to show that this limit"},{"Start":"02:34.730 ","End":"02:38.359","Text":"exists and is equal to L."},{"Start":"02:38.359 ","End":"02:45.060","Text":"But that\u0027s exactly what\u0027s given here."},{"Start":"02:45.060 ","End":"02:50.210","Text":"We\u0027re given that this limit is equal to L. Instead of x,"},{"Start":"02:50.210 ","End":"02:53.030","Text":"we have h but that\u0027s the same thing."},{"Start":"02:53.030 ","End":"02:56.075","Text":"So that\u0027s Part a done."},{"Start":"02:56.075 ","End":"02:58.085","Text":"Now what about Part b?"},{"Start":"02:58.085 ","End":"03:00.935","Text":"Well, exactly the same proof works"},{"Start":"03:00.935 ","End":"03:04.310","Text":"because if you look at the statement of L\u0027Hospital\u0027s rule,"},{"Start":"03:04.310 ","End":"03:05.910","Text":"that you pause and read this,"},{"Start":"03:05.910 ","End":"03:08.540","Text":"but the important thing is that when we"},{"Start":"03:08.540 ","End":"03:11.930","Text":"take the limit of the quotient of the derivatives,"},{"Start":"03:11.930 ","End":"03:16.925","Text":"we need for the limit on the right to either exist or be infinite."},{"Start":"03:16.925 ","End":"03:19.730","Text":"Everything works for the infinite case also,"},{"Start":"03:19.730 ","End":"03:27.350","Text":"if you just follow the solution in Part a and replace L by infinity everywhere it works,"},{"Start":"03:27.350 ","End":"03:34.050","Text":"because L\u0027Hospital\u0027s works for the infinite case. That\u0027s it."}],"ID":26174},{"Watched":false,"Name":"Exercise 8 - Dirichlet function","Duration":"3m 25s","ChapterTopicVideoID":25359,"CourseChapterTopicPlaylistID":257167,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25359.jpeg","UploadDate":"2021-04-01T16:50:37.7330000","DurationForVideoObject":"PT3M25S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.880","Text":"In this exercise, we\u0027re going to consider"},{"Start":"00:02.880 ","End":"00:06.645","Text":"a famous function called the Dirichlet function,"},{"Start":"00:06.645 ","End":"00:11.250","Text":"and it\u0027s defined as being"},{"Start":"00:11.250 ","End":"00:17.190","Text":"1 for rational numbers and 0 for irrational numbers."},{"Start":"00:17.190 ","End":"00:20.640","Text":"Sometimes it\u0027s defined the other way around."},{"Start":"00:20.640 ","End":"00:25.580","Text":"Question is, is it possible that the Dirichlet function is"},{"Start":"00:25.580 ","End":"00:31.430","Text":"the derivative of a differentiable function on 0, 1."},{"Start":"00:31.430 ","End":"00:33.920","Text":"I can\u0027t sketch this function."},{"Start":"00:33.920 ","End":"00:35.990","Text":"It\u0027s too crazy, but thing is,"},{"Start":"00:35.990 ","End":"00:38.875","Text":"it only has 2 values, 0 or 1."},{"Start":"00:38.875 ","End":"00:41.525","Text":"Now, turns out it\u0027s not possible."},{"Start":"00:41.525 ","End":"00:45.380","Text":"We\u0027ll do a proof by contradiction and we\u0027ll use Darboux\u0027s theorem."},{"Start":"00:45.380 ","End":"00:47.630","Text":"Suppose such an f exists,"},{"Start":"00:47.630 ","End":"00:53.000","Text":"then choose any irrational number between 0 and 1."},{"Start":"00:53.000 ","End":"00:55.145","Text":"You can take Pi over 4,"},{"Start":"00:55.145 ","End":"00:58.595","Text":"square root of 2/2, 1/ a,"},{"Start":"00:58.595 ","End":"01:02.490","Text":"root 2 minus 1, plenty of possibilities."},{"Start":"01:02.490 ","End":"01:04.060","Text":"Just choose 1."},{"Start":"01:04.060 ","End":"01:13.765","Text":"Then what we have is that f prime of 0 is equal to 1 because 0 is rational,"},{"Start":"01:13.765 ","End":"01:18.080","Text":"and f prime of a is 0 because a is irrational."},{"Start":"01:18.080 ","End":"01:26.995","Text":"So we chose it. Now we can apply Darboux\u0027s theorem to the interval from 0 to a."},{"Start":"01:26.995 ","End":"01:31.335","Text":"We\u0027ll take a number between this 1 and 0, choose 1/2,"},{"Start":"01:31.335 ","End":"01:36.110","Text":"which is between 0 and 1 or between 1 and 0 if you look at it backwards."},{"Start":"01:36.110 ","End":"01:38.525","Text":"So by the theorem,"},{"Start":"01:38.525 ","End":"01:42.950","Text":"there is some c in the interval from 0 to a,"},{"Start":"01:42.950 ","End":"01:49.355","Text":"not including the endpoints such that f prime of c is 1/2."},{"Start":"01:49.355 ","End":"01:51.545","Text":"How can f prime of c be 1/2?"},{"Start":"01:51.545 ","End":"01:54.220","Text":"It only has 2 values, this function,"},{"Start":"01:54.220 ","End":"01:58.520","Text":"D of c is 1/2, can\u0027t be because it only has 0s and 1s."},{"Start":"01:58.520 ","End":"02:00.650","Text":"So that\u0027s not possible."},{"Start":"02:00.650 ","End":"02:06.835","Text":"That basically concludes the proof of why the Dirichlet function is not a derivative."},{"Start":"02:06.835 ","End":"02:11.000","Text":"Remark, If f prime had had"},{"Start":"02:11.000 ","End":"02:16.730","Text":"a removable discontinuity or jump discontinuity or infinite discontinuity,"},{"Start":"02:16.730 ","End":"02:18.860","Text":"then by what we saw,"},{"Start":"02:18.860 ","End":"02:21.500","Text":"we could have said no right away."},{"Start":"02:21.500 ","End":"02:24.305","Text":"This is not the derivative of any function,"},{"Start":"02:24.305 ","End":"02:26.435","Text":"but in this case,"},{"Start":"02:26.435 ","End":"02:29.430","Text":"it\u0027s not 1 of those 3 types."},{"Start":"02:29.430 ","End":"02:31.820","Text":"It\u0027s an other type."},{"Start":"02:31.820 ","End":"02:33.680","Text":"It\u0027s part of type 2,"},{"Start":"02:33.680 ","End":"02:36.050","Text":"but type 2 also includes the infinite ones,"},{"Start":"02:36.050 ","End":"02:41.350","Text":"but this is the 1 where 1 of the 1-sided limits or both doesn\u0027t exist."},{"Start":"02:41.350 ","End":"02:44.430","Text":"In fact, this function has no points of"},{"Start":"02:44.430 ","End":"02:49.880","Text":"continuity because near every point there are rationals and irrationals,"},{"Start":"02:49.880 ","End":"02:53.150","Text":"so is the function going to be 0 or 1?"},{"Start":"02:53.150 ","End":"02:54.515","Text":"It\u0027s a crazy function."},{"Start":"02:54.515 ","End":"02:57.320","Text":"So what is this other type of discontinuity?"},{"Start":"02:57.320 ","End":"03:00.410","Text":"You can\u0027t actually say right away yes or no."},{"Start":"03:00.410 ","End":"03:02.345","Text":"There are examples of both."},{"Start":"03:02.345 ","End":"03:06.120","Text":"We had to use Darboux\u0027s theorem directly,"},{"Start":"03:06.120 ","End":"03:09.320","Text":"more creatively to figure out how to prove or disprove."},{"Start":"03:09.320 ","End":"03:12.470","Text":"In this case, we showed that it\u0027s not the case."},{"Start":"03:12.470 ","End":"03:15.530","Text":"But there are other examples of functions that have"},{"Start":"03:15.530 ","End":"03:21.230","Text":"this crazy kind of discontinuity where the limit doesn\u0027t exist and they are derivatives."},{"Start":"03:21.230 ","End":"03:25.470","Text":"Okay, that\u0027s all for this clip, we are done."}],"ID":26175}],"Thumbnail":null,"ID":257167}]

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