WEBVTT 00:00:00.000 --> 00:00:01.992 align:middle line:90% [SQUEAKING] 00:00:01.992 --> 00:00:03.486 align:middle line:90% [RUSTLING] 00:00:03.486 --> 00:00:05.976 align:middle line:90% [CLICKING] 00:00:05.976 --> 00:00:12.310 align:middle line:90% 00:00:12.310 --> 00:00:14.560 align:middle line:84% CASEY RODRIGUEZ: So let's continue with our discussion 00:00:14.560 --> 00:00:15.520 align:middle line:90% of the derivative. 00:00:15.520 --> 00:00:24.040 align:middle line:90% 00:00:24.040 --> 00:00:27.970 align:middle line:84% Let me recall that the derivative of a function, if it 00:00:27.970 --> 00:00:34.790 align:middle line:84% exists, which we denote by f prime of c, this 00:00:34.790 --> 00:00:39.530 align:middle line:84% is the limit as x goes to c of f of x 00:00:39.530 --> 00:00:45.260 align:middle line:90% minus f of c over x minus c. 00:00:45.260 --> 00:00:51.500 align:middle line:84% Last time, we looked at the relationship between continuity 00:00:51.500 --> 00:00:56.290 align:middle line:84% and differentiability and showed that differentiability implies 00:00:56.290 --> 00:01:00.340 align:middle line:84% continuity, but that having a derivative 00:01:00.340 --> 00:01:03.880 align:middle line:90% is quite a miraculous thing. 00:01:03.880 --> 00:01:05.920 align:middle line:84% Because there are continuous functions, 00:01:05.920 --> 00:01:08.050 align:middle line:90% which we actually constructed. 00:01:08.050 --> 00:01:09.910 align:middle line:84% We constructed a continuous function. 00:01:09.910 --> 00:01:11.500 align:middle line:90% You can generalize that example. 00:01:11.500 --> 00:01:15.010 align:middle line:90% 00:01:15.010 --> 00:01:17.070 align:middle line:84% So we constructed a continuous function 00:01:17.070 --> 00:01:20.230 align:middle line:90% that is differentiable nowhere. 00:01:20.230 --> 00:01:25.440 align:middle line:84% So in fact, differentiability is a much stronger condition 00:01:25.440 --> 00:01:26.370 align:middle line:90% than continuity. 00:01:26.370 --> 00:01:27.885 align:middle line:90% And it's something of a miracle. 00:01:27.885 --> 00:01:34.200 align:middle line:90% 00:01:34.200 --> 00:01:39.200 align:middle line:84% Just as a side note, if you go on to study complex analysis, 00:01:39.200 --> 00:01:41.170 align:middle line:84% which are functions of a complex variable 00:01:41.170 --> 00:01:42.940 align:middle line:84% rather than of a real variable, you also 00:01:42.940 --> 00:01:44.990 align:middle line:84% have the notion of a derivative there. 00:01:44.990 --> 00:01:50.500 align:middle line:84% And then, the derivative is much more miraculous 00:01:50.500 --> 00:01:54.460 align:middle line:84% than it is here in the setting of r. 00:01:54.460 --> 00:01:56.200 align:middle line:84% But the whole point is, the derivative 00:01:56.200 --> 00:01:59.050 align:middle line:90% is something of a miracle. 00:01:59.050 --> 00:02:03.130 align:middle line:84% So today, the goal is to prove some properties 00:02:03.130 --> 00:02:04.150 align:middle line:90% about the derivatives. 00:02:04.150 --> 00:02:06.792 align:middle line:90% Most of these you know. 00:02:06.792 --> 00:02:08.500 align:middle line:84% Maybe you didn't cover the proofs of them 00:02:08.500 --> 00:02:14.500 align:middle line:84% in calculus with the ultimate goal of proving the mean value 00:02:14.500 --> 00:02:23.080 align:middle line:84% theorem, which is, to me probably 00:02:23.080 --> 00:02:26.440 align:middle line:84% the most underappreciated but most important 00:02:26.440 --> 00:02:28.610 align:middle line:90% result in calculus. 00:02:28.610 --> 00:02:32.200 align:middle line:84% Now, you could argue that probably 00:02:32.200 --> 00:02:33.925 align:middle line:84% the hero of the story of calculus 00:02:33.925 --> 00:02:35.800 align:middle line:84% is the fundamental theorem of calculus, which 00:02:35.800 --> 00:02:38.140 align:middle line:84% connects integration, which we'll cover next 00:02:38.140 --> 00:02:39.980 align:middle line:90% after differentiation. 00:02:39.980 --> 00:02:42.310 align:middle line:84% Which so the fundamental theorem of calculus 00:02:42.310 --> 00:02:46.320 align:middle line:84% connects integration and differentiation. 00:02:46.320 --> 00:02:47.720 align:middle line:90% So that really is, but-- 00:02:47.720 --> 00:02:49.850 align:middle line:84% and that is really the hero of calculus. 00:02:49.850 --> 00:02:55.220 align:middle line:84% It gets used more than anything else. 00:02:55.220 --> 00:03:00.890 align:middle line:84% But to prove that, you use the mean value theorem. 00:03:00.890 --> 00:03:06.093 align:middle line:84% So if the fundamental theorem of calculus was Batman, 00:03:06.093 --> 00:03:07.510 align:middle line:84% I would say the mean value theorem 00:03:07.510 --> 00:03:09.460 align:middle line:90% is something like Alfred. 00:03:09.460 --> 00:03:12.730 align:middle line:84% He's really the reason why Batman gets to be who he is. 00:03:12.730 --> 00:03:16.060 align:middle line:90% 00:03:16.060 --> 00:03:17.170 align:middle line:90% Enough analogies for now. 00:03:17.170 --> 00:03:23.100 align:middle line:84% Let's go back to proving some properties of the derivative. 00:03:23.100 --> 00:03:30.902 align:middle line:90% So first, let's do some basic-- 00:03:30.902 --> 00:03:35.310 align:middle line:84% the basic linearity and rules for the derivative. 00:03:35.310 --> 00:03:43.505 align:middle line:84% So let f go from I to R. i is always some interval. 00:03:43.505 --> 00:03:53.310 align:middle line:90% 00:03:53.310 --> 00:03:57.990 align:middle line:84% And if f-- so I have two functions, f and g, 00:03:57.990 --> 00:04:00.382 align:middle line:90% from I to R. c is a point in i. 00:04:00.382 --> 00:04:03.350 align:middle line:90% 00:04:03.350 --> 00:04:17.200 align:middle line:84% And the conclusion is, if f and g are differentiable at c, 00:04:17.200 --> 00:04:22.570 align:middle line:84% then we have several rules that we can apply. 00:04:22.570 --> 00:04:31.730 align:middle line:84% The first is basically the linearity of the derivative 00:04:31.730 --> 00:04:38.870 align:middle line:84% that for all alpha and R, function alpha f plus 00:04:38.870 --> 00:04:49.910 align:middle line:84% g, which now goes from I to R, is differentiable at c. 00:04:49.910 --> 00:04:52.520 align:middle line:90% 00:04:52.520 --> 00:04:58.040 align:middle line:84% And the derivative of the function alpha f plus g 00:04:58.040 --> 00:05:04.520 align:middle line:84% prime c is alpha times f prime of c plus g prime of c. 00:05:04.520 --> 00:05:09.575 align:middle line:90% 00:05:09.575 --> 00:05:14.340 align:middle line:84% I think I'm starting to write a little slandered here. 00:05:14.340 --> 00:05:24.300 align:middle line:84% Second is the multiplication rule, which is the function 00:05:24.300 --> 00:05:34.460 align:middle line:84% f times g is differentiable at c. 00:05:34.460 --> 00:05:41.030 align:middle line:84% And the derivative of the product 00:05:41.030 --> 00:05:42.920 align:middle line:84% is not the product of the derivatives, 00:05:42.920 --> 00:05:50.270 align:middle line:84% but the prime of c, g of c plus f of c, g prime of c. 00:05:50.270 --> 00:05:53.160 align:middle line:90% 00:05:53.160 --> 00:05:55.740 align:middle line:90% So that's a multiplication rule. 00:05:55.740 --> 00:05:58.020 align:middle line:84% And then, we also have the quotient rule. 00:05:58.020 --> 00:06:10.130 align:middle line:90% 00:06:10.130 --> 00:06:13.040 align:middle line:84% And just like when you need to divide by something, 00:06:13.040 --> 00:06:17.270 align:middle line:84% you need to assume that you're never dividing by 0. 00:06:17.270 --> 00:06:24.560 align:middle line:84% If g of x does not equal 0 for all x in I, 00:06:24.560 --> 00:06:40.710 align:middle line:84% then the function f over g is differentiable at c. 00:06:40.710 --> 00:06:48.910 align:middle line:84% And the derivative is the derivative 00:06:48.910 --> 00:06:53.950 align:middle line:84% of the top times the bottom minus the top 00:06:53.950 --> 00:06:58.210 align:middle line:84% times the derivative of the bottom over the bottom 00:06:58.210 --> 00:07:04.450 align:middle line:84% squared, [? now ?] prime g of c squared. 00:07:04.450 --> 00:07:08.380 align:middle line:90% 00:07:08.380 --> 00:07:10.480 align:middle line:90% So I'll prove one and two. 00:07:10.480 --> 00:07:13.030 align:middle line:84% Three we'll leave as an exercise. 00:07:13.030 --> 00:07:27.550 align:middle line:90% 00:07:27.550 --> 00:07:33.700 align:middle line:84% So for one, we just compute that the limit 00:07:33.700 --> 00:07:45.070 align:middle line:84% as x goes to c of alpha f plus g of x, minus alpha f plus g of c 00:07:45.070 --> 00:07:53.040 align:middle line:84% over x minus c, this is just by definition alpha times 00:07:53.040 --> 00:07:57.990 align:middle line:84% f of x plus g of x minus alpha times f of c plus g of c. 00:07:57.990 --> 00:08:02.790 align:middle line:84% And so, this equals the limit as x goes to c. 00:08:02.790 --> 00:08:07.020 align:middle line:84% And just collecting terms alpha times alpha 00:08:07.020 --> 00:08:15.480 align:middle line:84% f of x minus f of c times x minus c, plus g of x 00:08:15.480 --> 00:08:20.280 align:middle line:90% minus g of c over x minus c. 00:08:20.280 --> 00:08:27.790 align:middle line:90% And all of these limits exist. 00:08:27.790 --> 00:08:29.230 align:middle line:90% So this is just a fixed number. 00:08:29.230 --> 00:08:33.100 align:middle line:84% The limit as x goes to c of this exists and this exists. 00:08:33.100 --> 00:08:36.419 align:middle line:84% So by what we know about limits, namely 00:08:36.419 --> 00:08:40.929 align:middle line:84% that the limit of the sum is the sum of the limits. 00:08:40.929 --> 00:08:43.090 align:middle line:84% And then, scalars just pull out of the limit. 00:08:43.090 --> 00:08:49.900 align:middle line:84% This is equal to alpha times f prime of c plus g prime of c. 00:08:49.900 --> 00:08:52.750 align:middle line:84% So that's the proof of the first one. 00:08:52.750 --> 00:08:56.590 align:middle line:84% For the proof of the second one, we'll 00:08:56.590 --> 00:09:00.490 align:middle line:84% use the fact that a function which 00:09:00.490 --> 00:09:06.110 align:middle line:84% is differentiable at a point is also continuous at a point. 00:09:06.110 --> 00:09:21.880 align:middle line:84% So since g is differentiable at c, it is continuous at c. 00:09:21.880 --> 00:09:23.350 align:middle line:90% I.e. 00:09:23.350 --> 00:09:29.350 align:middle line:84% limit as x goes to c of g of x equals g of c. 00:09:29.350 --> 00:09:32.650 align:middle line:90% 00:09:32.650 --> 00:09:34.930 align:middle line:84% Now we compute the limit as x goes 00:09:34.930 --> 00:09:42.310 align:middle line:84% to c of f times g of x minus f of c 00:09:42.310 --> 00:09:45.700 align:middle line:90% times g of c over x minus c. 00:09:45.700 --> 00:09:49.330 align:middle line:90% 00:09:49.330 --> 00:09:52.120 align:middle line:84% Now what I'm going to do is add and subtract 00:09:52.120 --> 00:09:54.790 align:middle line:90% f of c times g of x. 00:09:54.790 --> 00:09:57.340 align:middle line:84% So this I can write as the limit as x 00:09:57.340 --> 00:10:06.070 align:middle line:84% goes to c of f of x, minus f of c over x minus c, times g of x. 00:10:06.070 --> 00:10:10.380 align:middle line:90% 00:10:10.380 --> 00:10:22.640 align:middle line:84% Plus now, f of c times g of x minus g of c over x minus c. 00:10:22.640 --> 00:10:26.230 align:middle line:90% 00:10:26.230 --> 00:10:30.250 align:middle line:84% And again, so this limit as x goes to c exists. 00:10:30.250 --> 00:10:34.090 align:middle line:84% This is just the derivative of f at c. 00:10:34.090 --> 00:10:36.655 align:middle line:84% And we noted that since g is differentiable at c, 00:10:36.655 --> 00:10:39.130 align:middle line:84% g is also continuous at c, meaning the limit 00:10:39.130 --> 00:10:42.280 align:middle line:84% as x goes to c of g of x equals g of c. 00:10:42.280 --> 00:10:43.900 align:middle line:90% This is just a constant f of c. 00:10:43.900 --> 00:10:45.610 align:middle line:90% This goes to g prime of c. 00:10:45.610 --> 00:10:48.610 align:middle line:90% So all these limits exist. 00:10:48.610 --> 00:10:50.410 align:middle line:84% And therefore, we get that this is 00:10:50.410 --> 00:10:59.500 align:middle line:84% equal to f prime of c times g of c plus f of c, g prime of c. 00:10:59.500 --> 00:11:01.540 align:middle line:84% And three, I'll leave as an exercise. 00:11:01.540 --> 00:11:05.450 align:middle line:90% 00:11:05.450 --> 00:11:07.220 align:middle line:90% Again, you'll write out-- 00:11:07.220 --> 00:11:10.220 align:middle line:90% 00:11:10.220 --> 00:11:12.110 align:middle line:84% you'll add and subtract something 00:11:12.110 --> 00:11:16.550 align:middle line:84% to f of x over g of x minus f of c over g of c 00:11:16.550 --> 00:11:20.090 align:middle line:84% and use the fact that differentiability at a point 00:11:20.090 --> 00:11:22.970 align:middle line:84% implies continuity, and just evaluate the limits. 00:11:22.970 --> 00:11:26.030 align:middle line:84% So I'm going to stop with these two rules. 00:11:26.030 --> 00:11:32.412 align:middle line:90% 00:11:32.412 --> 00:11:34.620 align:middle line:84% I don't know why I called it the multiplication rule. 00:11:34.620 --> 00:11:36.230 align:middle line:90% It's called the product rule-- 00:11:36.230 --> 00:11:39.020 align:middle line:84% well, multiplication and product are the same thing. 00:11:39.020 --> 00:11:45.560 align:middle line:90% 00:11:45.560 --> 00:11:47.800 align:middle line:84% But we also have the chain rule, which 00:11:47.800 --> 00:11:54.340 align:middle line:84% requires a little more care to prove than what we've 00:11:54.340 --> 00:11:57.310 align:middle line:90% done so far for these rules. 00:11:57.310 --> 00:12:03.550 align:middle line:90% So suppose I have two intervals. 00:12:03.550 --> 00:12:09.730 align:middle line:90% 00:12:09.730 --> 00:12:15.130 align:middle line:84% And then I have a function g going from I1 to I2, 00:12:15.130 --> 00:12:25.740 align:middle line:84% f going from I2 to R. And suppose g 00:12:25.740 --> 00:12:39.160 align:middle line:84% is differentiable at c, and f is differentiable at g of c. 00:12:39.160 --> 00:12:44.850 align:middle line:90% 00:12:44.850 --> 00:12:58.800 align:middle line:84% Then the function f of g, function goes from I1 to R. 00:12:58.800 --> 00:13:02.850 align:middle line:90% This is differentiable at c. 00:13:02.850 --> 00:13:16.330 align:middle line:84% And the derivative equals f prime of g of c 00:13:16.330 --> 00:13:19.030 align:middle line:90% times g prime of c. 00:13:19.030 --> 00:13:38.270 align:middle line:90% 00:13:38.270 --> 00:13:51.170 align:middle line:90% So what's the basic idea? 00:13:51.170 --> 00:13:58.310 align:middle line:84% Again, so this is basic idea is that we write the difference 00:13:58.310 --> 00:14:11.530 align:middle line:84% quotient, g of x minus f of g of c over x minus c as f of g of x 00:14:11.530 --> 00:14:19.460 align:middle line:84% minus g, f at g of c times over g 00:14:19.460 --> 00:14:31.080 align:middle line:84% of x minus g of c times g of x over g of c times x minus c. 00:14:31.080 --> 00:14:35.190 align:middle line:84% And let x go to c, and then we'll pick up 00:14:35.190 --> 00:14:37.200 align:middle line:90% the derivative of g here at c. 00:14:37.200 --> 00:14:41.970 align:middle line:84% And right here, this is f of something converging to g of c. 00:14:41.970 --> 00:14:45.150 align:middle line:84% And so, this should be like f prime of g of c. 00:14:45.150 --> 00:14:48.120 align:middle line:84% But the only problem with really writing this down 00:14:48.120 --> 00:14:51.720 align:middle line:84% is that there could be points where g of x equals g of c. 00:14:51.720 --> 00:14:53.040 align:middle line:90% And then, I'm dividing by 0. 00:14:53.040 --> 00:14:54.210 align:middle line:90% That's not allowed. 00:14:54.210 --> 00:14:59.310 align:middle line:84% So writing this expression is meaningless. 00:14:59.310 --> 00:15:02.250 align:middle line:84% But we're going to do something like this in spirit, 00:15:02.250 --> 00:15:06.360 align:middle line:84% where at least when g of x does not equal g of c, 00:15:06.360 --> 00:15:08.430 align:middle line:84% this thing kind of equals this thing. 00:15:08.430 --> 00:15:11.690 align:middle line:90% 00:15:11.690 --> 00:15:14.945 align:middle line:90% That's the basic idea. 00:15:14.945 --> 00:15:19.640 align:middle line:90% 00:15:19.640 --> 00:15:27.420 align:middle line:84% So to implement this strategy, let's 00:15:27.420 --> 00:15:31.810 align:middle line:84% introduce some notation and a few auxiliary functions. 00:15:31.810 --> 00:15:38.760 align:middle line:84% So first off, the function we're interested in taking 00:15:38.760 --> 00:15:41.040 align:middle line:84% the derivative f of g, let me call that h of x. 00:15:41.040 --> 00:15:47.610 align:middle line:84% Because I don't want to have to keep writing f of g of x. 00:15:47.610 --> 00:15:48.720 align:middle line:90% Let's call it d. 00:15:48.720 --> 00:15:52.340 align:middle line:90% This will be the point g of c. 00:15:52.340 --> 00:16:02.610 align:middle line:84% So we want to show then that h prime of c 00:16:02.610 --> 00:16:08.460 align:middle line:84% exists, and it's equal to f prime of d times g prime of c. 00:16:08.460 --> 00:16:12.180 align:middle line:90% 00:16:12.180 --> 00:16:14.970 align:middle line:84% Just in this notation I've set up. 00:16:14.970 --> 00:16:23.600 align:middle line:84% So now, let me define some auxiliary functions u of y. 00:16:23.600 --> 00:16:29.030 align:middle line:84% This is going to be essentially the difference quotient of f, 00:16:29.030 --> 00:16:30.920 align:middle line:90% except at a certain point. 00:16:30.920 --> 00:16:35.990 align:middle line:84% So this is f of y minus f of d over y 00:16:35.990 --> 00:16:42.390 align:middle line:84% minus d when y does not equal d so that this is meaningful. 00:16:42.390 --> 00:16:47.690 align:middle line:84% And then, f prime of d when y equals d. 00:16:47.690 --> 00:16:50.840 align:middle line:90% And then, v of x-- 00:16:50.840 --> 00:16:55.580 align:middle line:84% so this is going to play the role of that first expression 00:16:55.580 --> 00:16:58.070 align:middle line:84% I wrote up there, f of g of x minus f of g of c. 00:16:58.070 --> 00:17:01.790 align:middle line:90% 00:17:01.790 --> 00:17:04.760 align:middle line:84% And then, v of x, this will be g of x 00:17:04.760 --> 00:17:12.170 align:middle line:84% minus g of c, x minus c when x does not equal c, 00:17:12.170 --> 00:17:16.460 align:middle line:84% and g prime of c when x equals c. 00:17:16.460 --> 00:17:19.670 align:middle line:90% 00:17:19.670 --> 00:17:33.000 align:middle line:84% Now, essentially what do we have here? 00:17:33.000 --> 00:17:36.740 align:middle line:90% 00:17:36.740 --> 00:17:42.030 align:middle line:84% Then if I look at f of y minus f of d, 00:17:42.030 --> 00:17:49.020 align:middle line:84% this is equal to u of y times y minus d, always. 00:17:49.020 --> 00:17:53.640 align:middle line:84% Because if y does not equal d, then I divide over, 00:17:53.640 --> 00:17:57.240 align:middle line:84% and I get u of y, then I get f of y 00:17:57.240 --> 00:17:59.790 align:middle line:84% minus f of d on the left-hand side over y minus d. 00:17:59.790 --> 00:18:02.440 align:middle line:84% And that's equal to u of y by definition. 00:18:02.440 --> 00:18:06.900 align:middle line:84% And then, when y equals d, I get 0 here, and I get 0 over here. 00:18:06.900 --> 00:18:09.060 align:middle line:90% So that's clear. 00:18:09.060 --> 00:18:14.910 align:middle line:84% And also, the same here, g of x minus g of c 00:18:14.910 --> 00:18:19.590 align:middle line:84% equals v of x minus [INAUDIBLE] times x minus c. 00:18:19.590 --> 00:18:27.840 align:middle line:90% 00:18:27.840 --> 00:18:32.075 align:middle line:84% So we have these two expressions. 00:18:32.075 --> 00:18:33.450 align:middle line:84% And one more thing I want to note 00:18:33.450 --> 00:18:36.570 align:middle line:84% is that u and v are continuous at y equals 00:18:36.570 --> 00:18:39.690 align:middle line:90% d and x equals c respectively. 00:18:39.690 --> 00:18:42.280 align:middle line:90% 00:18:42.280 --> 00:18:53.230 align:middle line:84% So note that v of y is continuous at d, 00:18:53.230 --> 00:18:59.050 align:middle line:90% and v of x is continuous at c. 00:18:59.050 --> 00:19:01.810 align:middle line:84% So let's look at the proof of that. 00:19:01.810 --> 00:19:06.330 align:middle line:84% We just need to show that the limit as y goes to d of v of y 00:19:06.330 --> 00:19:07.710 align:middle line:90% equals v of d. 00:19:07.710 --> 00:19:12.480 align:middle line:84% So compute the limit as y goes to d of v of y. 00:19:12.480 --> 00:19:16.740 align:middle line:84% Now, when we look at limits, remember, 00:19:16.740 --> 00:19:19.590 align:middle line:84% we're not allowed to put y equals 00:19:19.590 --> 00:19:23.280 align:middle line:84% d in the expression for this limit. 00:19:23.280 --> 00:19:27.000 align:middle line:84% We're always looking at points y close to d but not equal to d, 00:19:27.000 --> 00:19:28.810 align:middle line:90% is what I'm saying. 00:19:28.810 --> 00:19:32.670 align:middle line:84% And so, therefore, when y is near d, and not equal to d, 00:19:32.670 --> 00:19:35.910 align:middle line:90% v of y is given by-- 00:19:35.910 --> 00:19:37.081 align:middle line:90% I should-- this should-- 00:19:37.081 --> 00:19:40.240 align:middle line:90% 00:19:40.240 --> 00:19:42.110 align:middle line:84% I should have written u here, sorry, u. 00:19:42.110 --> 00:19:44.970 align:middle line:90% 00:19:44.970 --> 00:19:50.250 align:middle line:84% So we're looking at y near d, but not equal to d. 00:19:50.250 --> 00:19:53.220 align:middle line:84% And therefore, it's given by this expression. 00:19:53.220 --> 00:19:55.590 align:middle line:84% So this is the limit as y goes to d 00:19:55.590 --> 00:20:00.420 align:middle line:84% of f of y minus f of d over y minus d. 00:20:00.420 --> 00:20:05.820 align:middle line:84% And this is just the definition of the derivative of f at d. 00:20:05.820 --> 00:20:07.290 align:middle line:90% So remember, d is g of c. 00:20:07.290 --> 00:20:08.970 align:middle line:90% So f is differentiable there. 00:20:08.970 --> 00:20:14.340 align:middle line:84% And this is by definition u of d. 00:20:14.340 --> 00:20:20.790 align:middle line:84% So this function u is continuous at y equals d. 00:20:20.790 --> 00:20:30.540 align:middle line:84% And similarly, the limit as x goes to c of u of x, v of x 00:20:30.540 --> 00:20:37.940 align:middle line:84% equals v of c, which is g prime of c. 00:20:37.940 --> 00:20:41.000 align:middle line:90% 00:20:41.000 --> 00:20:46.250 align:middle line:84% So now we're going to combine what I've written here 00:20:46.250 --> 00:20:48.110 align:middle line:90% to finish the proof. 00:20:48.110 --> 00:21:00.560 align:middle line:84% So in h of x minus h of c, this is 00:21:00.560 --> 00:21:08.620 align:middle line:84% equal to f of g of x minus f of g of c. 00:21:08.620 --> 00:21:20.650 align:middle line:84% And this is equal to u of g of x times g of x minus g of c. 00:21:20.650 --> 00:21:24.650 align:middle line:90% This is from this relation here. 00:21:24.650 --> 00:21:31.940 align:middle line:84% So we'll call this one star, this one double star. 00:21:31.940 --> 00:21:32.705 align:middle line:90% This is by star. 00:21:32.705 --> 00:21:37.180 align:middle line:90% 00:21:37.180 --> 00:21:45.640 align:middle line:84% And now, I use the second relation, double star, 00:21:45.640 --> 00:21:55.330 align:middle line:84% to write g of x minus g of c equals v of x times x minus c. 00:21:55.330 --> 00:22:00.030 align:middle line:84% And therefore, the limit as x goes 00:22:00.030 --> 00:22:11.960 align:middle line:84% to c of h of x minus h of c, or x minus c, this 00:22:11.960 --> 00:22:16.130 align:middle line:84% is equal to the limit as x goes to c 00:22:16.130 --> 00:22:21.350 align:middle line:90% of u of g of x times v of x. 00:22:21.350 --> 00:22:28.700 align:middle line:84% Now, u is continuous at g of c, g is continuous at c. 00:22:28.700 --> 00:22:30.620 align:middle line:84% And we proved that the composition 00:22:30.620 --> 00:22:33.680 align:middle line:84% of two continuous functions is continuous. 00:22:33.680 --> 00:22:41.720 align:middle line:90% So this equals u of g of c. 00:22:41.720 --> 00:22:43.850 align:middle line:84% And v of x, we've proven already. 00:22:43.850 --> 00:22:47.060 align:middle line:90% That limit is v of c. 00:22:47.060 --> 00:22:51.020 align:middle line:84% And u of g of c, so this is d, remember, 00:22:51.020 --> 00:22:52.940 align:middle line:90% in the notation we had earlier. 00:22:52.940 --> 00:22:59.400 align:middle line:84% And therefore, that equals f prime of g of c. 00:22:59.400 --> 00:23:06.910 align:middle line:84% Remember, g of c was d, times g prime of c. 00:23:06.910 --> 00:23:12.300 align:middle line:90% 00:23:12.300 --> 00:23:14.680 align:middle line:84% So to go from this point to this point, 00:23:14.680 --> 00:23:16.920 align:middle line:84% we use the fact that u is continuous at c, 00:23:16.920 --> 00:23:19.020 align:middle line:90% which we proved. 00:23:19.020 --> 00:23:23.320 align:middle line:84% u is continuous at g of c, which we proved over there. 00:23:23.320 --> 00:23:27.633 align:middle line:90% And that g is continuous at c. 00:23:27.633 --> 00:23:30.050 align:middle line:84% And therefore, the composition of two continuous functions 00:23:30.050 --> 00:23:35.572 align:middle line:84% is continuous at that point to arrive at this guy. 00:23:35.572 --> 00:23:37.030 align:middle line:84% But again, the whole basic idea was 00:23:37.030 --> 00:23:38.560 align:middle line:84% to be able to write the difference 00:23:38.560 --> 00:23:44.290 align:middle line:84% quotient as a product of what looks like the difference 00:23:44.290 --> 00:23:48.050 align:middle line:84% quotient of f times the difference quotient of g. 00:23:48.050 --> 00:23:51.050 align:middle line:84% But you have to be careful, because g of x 00:23:51.050 --> 00:23:53.210 align:middle line:84% could equal g of c, in which case 00:23:53.210 --> 00:23:57.240 align:middle line:84% you're dividing by 0 in this basic idea we had up there, 00:23:57.240 --> 00:23:58.940 align:middle line:90% which is not allowed. 00:23:58.940 --> 00:24:04.670 align:middle line:84% But you get around it by using these functions which 00:24:04.670 --> 00:24:08.270 align:middle line:84% equal what you want as long as you're not dividing by 0, 00:24:08.270 --> 00:24:09.020 align:middle line:90% basically. 00:24:09.020 --> 00:24:14.990 align:middle line:90% 00:24:14.990 --> 00:24:21.411 align:middle line:84% So that wraps it up for basic properties of the derivative. 00:24:21.411 --> 00:24:24.120 align:middle line:90% 00:24:24.120 --> 00:24:32.960 align:middle line:84% And now, we're going to move on to trying to prove 00:24:32.960 --> 00:24:34.040 align:middle line:90% the mean value theorem-- 00:24:34.040 --> 00:24:45.990 align:middle line:90% 00:24:45.990 --> 00:24:48.855 align:middle line:90% the Alfred of this Batman tale. 00:24:48.855 --> 00:24:54.780 align:middle line:90% 00:24:54.780 --> 00:24:56.880 align:middle line:84% But first, in order to prove this, 00:24:56.880 --> 00:25:01.420 align:middle line:84% we'll need a result about the derivative in relation 00:25:01.420 --> 00:25:05.680 align:middle line:84% to certain relative minimum and relative maximum. 00:25:05.680 --> 00:25:08.125 align:middle line:84% So let me first define what this means. 00:25:08.125 --> 00:25:13.360 align:middle line:90% 00:25:13.360 --> 00:25:20.750 align:middle line:84% Let S be a subset of R, f be a function from S to R. 00:25:20.750 --> 00:25:35.700 align:middle line:84% We say f has a relative max at c in S. 00:25:35.700 --> 00:25:40.490 align:middle line:84% If there exists a delta positive, 00:25:40.490 --> 00:25:47.550 align:middle line:84% such that for all x in S, x minus c is less than delta, 00:25:47.550 --> 00:25:50.570 align:middle line:84% then f of x is less than or equal to f of c. 00:25:50.570 --> 00:25:54.260 align:middle line:90% 00:25:54.260 --> 00:25:56.870 align:middle line:84% So the picture here, and then let 00:25:56.870 --> 00:26:12.420 align:middle line:84% me state that there's analogous definition 00:26:12.420 --> 00:26:17.880 align:middle line:90% of relative min at c. 00:26:17.880 --> 00:26:19.500 align:middle line:84% You just take this entire definition 00:26:19.500 --> 00:26:22.560 align:middle line:90% and flip the inequality. 00:26:22.560 --> 00:26:24.210 align:middle line:84% That's the definition for relative min. 00:26:24.210 --> 00:26:27.660 align:middle line:84% So what this means pictorially, so remember, 00:26:27.660 --> 00:26:30.600 align:middle line:84% we had this notion of absolute min and absolute max. 00:26:30.600 --> 00:26:36.480 align:middle line:84% An absolute max at a point c means at every other point 00:26:36.480 --> 00:26:39.750 align:middle line:90% in S, f of x lies below f of c. 00:26:39.750 --> 00:26:45.390 align:middle line:84% Here we're just saying, as long as you're close to the point c, 00:26:45.390 --> 00:26:47.450 align:middle line:90% then you lay below f of c. 00:26:47.450 --> 00:27:05.430 align:middle line:84% So let's say our function looks something like that. 00:27:05.430 --> 00:27:08.900 align:middle line:90% 00:27:08.900 --> 00:27:21.040 align:middle line:84% So then, let's say we're at this point here. 00:27:21.040 --> 00:27:28.360 align:middle line:84% So this would be absolute max, because the graph 00:27:28.360 --> 00:27:34.600 align:middle line:84% of the function sits below the value of the function 00:27:34.600 --> 00:27:39.160 align:middle line:84% at this point for all x in this interval a, b. 00:27:39.160 --> 00:27:46.720 align:middle line:84% And then, here at this point, we would have an absolute min, 00:27:46.720 --> 00:27:49.630 align:middle line:90% because the rest of-- 00:27:49.630 --> 00:27:51.820 align:middle line:84% because the graph of the function 00:27:51.820 --> 00:27:58.050 align:middle line:84% sits above that point for all x in this interval. 00:27:58.050 --> 00:28:06.090 align:middle line:84% But now, if I go to this point, say, 00:28:06.090 --> 00:28:12.320 align:middle line:84% this would be a relative max, because why? 00:28:12.320 --> 00:28:18.290 align:middle line:84% If I look at an interval-- small interval around it, 00:28:18.290 --> 00:28:20.120 align:middle line:84% and I look inside this interval-- 00:28:20.120 --> 00:28:22.200 align:middle line:84% there's too many dashed lines going on. 00:28:22.200 --> 00:28:24.080 align:middle line:90% So I'm going to erase this. 00:28:24.080 --> 00:28:32.310 align:middle line:90% 00:28:32.310 --> 00:28:34.530 align:middle line:84% Yeah, I'm just going to do away with the dash. 00:28:34.530 --> 00:28:37.500 align:middle line:84% If I zoom in on just this one piece of the graph, 00:28:37.500 --> 00:28:44.260 align:middle line:84% then at this point, the graph sits 00:28:44.260 --> 00:28:55.000 align:middle line:84% below this point inside this little vertical strip. 00:28:55.000 --> 00:28:57.520 align:middle line:84% So this guy would be a relative max. 00:28:57.520 --> 00:29:00.550 align:middle line:84% And then similarly, at this point here, 00:29:00.550 --> 00:29:01.690 align:middle line:90% maybe it's-- let me-- 00:29:01.690 --> 00:29:04.580 align:middle line:90% 00:29:04.580 --> 00:29:14.990 align:middle line:84% so you can really see what the graph is. 00:29:14.990 --> 00:29:18.300 align:middle line:84% I should have brought different colors. 00:29:18.300 --> 00:29:22.500 align:middle line:84% At this point, we have a relative min. 00:29:22.500 --> 00:29:27.960 align:middle line:84% Because nearby points sit above the value of the function 00:29:27.960 --> 00:29:29.760 align:middle line:90% evaluated at this point. 00:29:29.760 --> 00:29:36.780 align:middle line:84% So relative max means nearby points f sits below 00:29:36.780 --> 00:29:39.360 align:middle line:90% that f evaluated at that point. 00:29:39.360 --> 00:29:41.190 align:middle line:84% And relative min is the other way around. 00:29:41.190 --> 00:29:49.550 align:middle line:90% 00:29:49.550 --> 00:29:53.600 align:middle line:84% So I don't think we need to belabor 00:29:53.600 --> 00:29:56.333 align:middle line:84% the point of what relative min relative maxes are. 00:29:56.333 --> 00:29:57.500 align:middle line:90% I think that's pretty clear. 00:29:57.500 --> 00:30:06.990 align:middle line:90% 00:30:06.990 --> 00:30:12.550 align:middle line:90% So the theorem is the following. 00:30:12.550 --> 00:30:17.840 align:middle line:90% 00:30:17.840 --> 00:30:29.970 align:middle line:84% If f from a, b has a relative min or max at a point 00:30:29.970 --> 00:30:34.300 align:middle line:84% c strictly inside this closed interval, in the open interval 00:30:34.300 --> 00:30:52.340 align:middle line:84% a, b, then and f is differentiable at c, then 00:30:52.340 --> 00:30:54.470 align:middle line:90% f prime of c equals 0. 00:30:54.470 --> 00:30:56.330 align:middle line:84% And this is kind of clear from this picture, 00:30:56.330 --> 00:30:58.880 align:middle line:84% although this might look more like a sawtooth, 00:30:58.880 --> 00:31:01.610 align:middle line:84% but it shouldn't, that the tangent at this point 00:31:01.610 --> 00:31:03.260 align:middle line:90% is horizontal. 00:31:03.260 --> 00:31:05.400 align:middle line:84% And the tangent at this point is horizontal, 00:31:05.400 --> 00:31:09.770 align:middle line:84% which just is the expression that we're just expressing 00:31:09.770 --> 00:31:11.840 align:middle line:90% that f prime of c equals 0. 00:31:11.840 --> 00:31:14.840 align:middle line:84% Because f prime of c is supposed to be the slope of the tangent. 00:31:14.840 --> 00:31:27.850 align:middle line:90% 00:31:27.850 --> 00:31:31.240 align:middle line:90% So the idea of the proof is-- 00:31:31.240 --> 00:31:33.070 align:middle line:84% and we'll do it for relative max. 00:31:33.070 --> 00:31:36.820 align:middle line:84% Relative min you get by taking minus f. 00:31:36.820 --> 00:31:52.520 align:middle line:84% So we'll do it for suppose f has a relative max at c in a, b. 00:31:52.520 --> 00:31:59.040 align:middle line:90% 00:31:59.040 --> 00:32:04.440 align:middle line:84% Then so I'm just going to draw a picture 00:32:04.440 --> 00:32:07.410 align:middle line:84% to illustrate why this is the case, so 00:32:07.410 --> 00:32:09.610 align:middle line:84% or why what I'm about to say is the case. 00:32:09.610 --> 00:32:13.650 align:middle line:90% So we have a, b, we have c. 00:32:13.650 --> 00:32:24.220 align:middle line:84% Then there exist delta positive, say delta one, 00:32:24.220 --> 00:32:26.830 align:middle line:90% such that two things are true. 00:32:26.830 --> 00:32:32.730 align:middle line:90% 00:32:32.730 --> 00:32:42.510 align:middle line:84% x or c minus delta 1 c plus delta 1 is contained in a, b. 00:32:42.510 --> 00:32:54.460 align:middle line:84% Two, for all x in c minus delta 1, c plus delta 1, 00:32:54.460 --> 00:32:57.790 align:middle line:84% f of x is less than or equal to f of c. 00:32:57.790 --> 00:33:04.120 align:middle line:84% So delta 1 is chosen as the minimum of-- 00:33:04.120 --> 00:33:06.757 align:middle line:84% so maybe let me not call it delta one, 00:33:06.757 --> 00:33:07.840 align:middle line:90% let me just call it delta. 00:33:07.840 --> 00:33:11.060 align:middle line:90% 00:33:11.060 --> 00:33:16.130 align:middle line:84% So first off, you know that a, b, open interval, 00:33:16.130 --> 00:33:20.490 align:middle line:84% this is an open set which we've encountered in the assignments. 00:33:20.490 --> 00:33:34.570 align:middle line:84% So you can find a delta so that delta 0, 00:33:34.570 --> 00:33:37.480 align:middle line:84% say, so that this interval centered 00:33:37.480 --> 00:33:39.310 align:middle line:90% at c is contained in a, b. 00:33:39.310 --> 00:33:41.980 align:middle line:90% 00:33:41.980 --> 00:33:46.540 align:middle line:84% Now, since f has a relative maximum at this point, 00:33:46.540 --> 00:33:49.090 align:middle line:84% there exist another delta 1, say, 00:33:49.090 --> 00:33:57.710 align:middle line:84% so that if I'm inside that interval, 00:33:57.710 --> 00:34:00.470 align:middle line:84% then f of x is less than or equal to f of c 00:34:00.470 --> 00:34:03.140 align:middle line:90% for all x in that interval. 00:34:03.140 --> 00:34:07.190 align:middle line:84% And I choose delta to be the minimum of these two deltas. 00:34:07.190 --> 00:34:12.370 align:middle line:84% And then I will have a chosen delta 00:34:12.370 --> 00:34:16.435 align:middle line:84% so that c minus delta c plus delta is-- 00:34:16.435 --> 00:34:21.360 align:middle line:90% 00:34:21.360 --> 00:34:26.264 align:middle line:90% so here, this one is-- 00:34:26.264 --> 00:34:34.840 align:middle line:90% 00:34:34.840 --> 00:34:37.400 align:middle line:84% and then delta 0 shows them so that that is contained. 00:34:37.400 --> 00:34:39.130 align:middle line:84% So I choose the minimum of these deltas. 00:34:39.130 --> 00:34:41.650 align:middle line:84% And this is the delta I take for this statement. 00:34:41.650 --> 00:34:45.600 align:middle line:90% 00:34:45.600 --> 00:34:50.880 align:middle line:84% So now, the way we get that the derivative is 0 00:34:50.880 --> 00:34:57.090 align:middle line:84% is, we're going to approach c from above and from 00:34:57.090 --> 00:35:01.920 align:middle line:84% below and use the fact that we have a relative max 00:35:01.920 --> 00:35:05.850 align:middle line:84% to show that the derivative is both bigger than or equal to 0 00:35:05.850 --> 00:35:07.920 align:middle line:90% and less than or equal to 0. 00:35:07.920 --> 00:35:17.840 align:middle line:90% So let xn be c minus-- 00:35:17.840 --> 00:35:19.340 align:middle line:90% how did I write this-- 00:35:19.340 --> 00:35:27.890 align:middle line:84% delta over 2n, which is in the interval c minus delta c 00:35:27.890 --> 00:35:28.730 align:middle line:90% for all n. 00:35:28.730 --> 00:35:33.940 align:middle line:90% 00:35:33.940 --> 00:35:38.290 align:middle line:84% Then what do we know? xn, the limit as n 00:35:38.290 --> 00:35:44.155 align:middle line:84% goes to infinity of c minus delta over 2, n converges to c. 00:35:44.155 --> 00:35:47.030 align:middle line:90% 00:35:47.030 --> 00:35:53.750 align:middle line:84% So the derivative at c is equal to the limit 00:35:53.750 --> 00:35:59.090 align:middle line:84% as n goes to infinity of f of x sub n minus f 00:35:59.090 --> 00:36:02.720 align:middle line:90% of c over x sub n minus c. 00:36:02.720 --> 00:36:07.640 align:middle line:90% Now, what do we note? 00:36:07.640 --> 00:36:10.170 align:middle line:84% So x sub is in this interval here, 00:36:10.170 --> 00:36:12.470 align:middle line:84% which is contained in this interval here. 00:36:12.470 --> 00:36:14.600 align:middle line:84% And therefore, f of x sub n is always 00:36:14.600 --> 00:36:16.370 align:middle line:90% less than or equal to f of c. 00:36:16.370 --> 00:36:20.915 align:middle line:84% So what's on top is always less than or equal to 0. 00:36:20.915 --> 00:36:23.890 align:middle line:90% 00:36:23.890 --> 00:36:26.770 align:middle line:84% Now, x sub n is always less than c. 00:36:26.770 --> 00:36:31.120 align:middle line:84% So this thing on bottom is also less than 0. 00:36:31.120 --> 00:36:33.010 align:middle line:84% And therefore, something that's less than 00:36:33.010 --> 00:36:35.770 align:middle line:84% or equal to 0 on top over something that's less than 0 00:36:35.770 --> 00:36:39.010 align:middle line:84% on bottom must be bigger than or equal to 0. 00:36:39.010 --> 00:36:45.660 align:middle line:90% 00:36:45.660 --> 00:36:48.510 align:middle line:84% So that's approaching c from the left. 00:36:48.510 --> 00:36:55.200 align:middle line:84% If we approach c from the right, delta over 2n 00:36:55.200 --> 00:37:05.190 align:middle line:84% which is contained in c, c plus delta for all n, then 00:37:05.190 --> 00:37:09.030 align:middle line:90% xn converges to c. 00:37:09.030 --> 00:37:14.070 align:middle line:84% And if we look at f prime of c, again, this 00:37:14.070 --> 00:37:17.550 align:middle line:84% is equal to limit as n goes to infinity 00:37:17.550 --> 00:37:23.040 align:middle line:84% of f of y sub n minus f of c over y sub n minus c. 00:37:23.040 --> 00:37:26.820 align:middle line:84% Again, f of c is a relative maximum. 00:37:26.820 --> 00:37:30.870 align:middle line:84% So this thing on top is still less than or equal to 0. 00:37:30.870 --> 00:37:32.760 align:middle line:84% And the thing on bottom though, is now 00:37:32.760 --> 00:37:37.140 align:middle line:84% since y sub n is bigger than c, this is positive. 00:37:37.140 --> 00:37:40.710 align:middle line:84% So something on top is negative, something on top is positive. 00:37:40.710 --> 00:37:43.290 align:middle line:90% So the limit must be negative. 00:37:43.290 --> 00:37:47.050 align:middle line:90% 00:37:47.050 --> 00:37:49.540 align:middle line:84% So we've just shown f prime of c is 00:37:49.540 --> 00:37:51.340 align:middle line:90% both bigger than or equal to 0. 00:37:51.340 --> 00:37:53.320 align:middle line:84% And that it's also less than or equal to 0. 00:37:53.320 --> 00:38:03.500 align:middle line:90% 00:38:03.500 --> 00:38:07.170 align:middle line:84% And therefore, the derivative is 0 at this point. 00:38:07.170 --> 00:38:13.295 align:middle line:90% 00:38:13.295 --> 00:38:15.170 align:middle line:84% And again, the proof for a relative minimum 00:38:15.170 --> 00:38:28.893 align:middle line:84% is very similar, except that inequality flips signs. 00:38:28.893 --> 00:38:30.810 align:middle line:84% And therefore, you would get this is less than 00:38:30.810 --> 00:38:34.530 align:middle line:84% or equal to 0, this one would be bigger than or equal to 0. 00:38:34.530 --> 00:38:38.400 align:middle line:84% But you still get f prime of c is bounded between 0 and 0. 00:38:38.400 --> 00:38:41.250 align:middle line:90% 00:38:41.250 --> 00:38:49.680 align:middle line:84% So at a point where we have a relative minimum strictly 00:38:49.680 --> 00:38:55.450 align:middle line:84% inside the interval, the derivative must be 0. 00:38:55.450 --> 00:39:00.915 align:middle line:90% 00:39:00.915 --> 00:39:03.290 align:middle line:84% And the fact that it happens strictly inside the interval 00:39:03.290 --> 00:39:04.100 align:middle line:90% is important. 00:39:04.100 --> 00:39:09.050 align:middle line:84% This is not necessarily true if the relative max 00:39:09.050 --> 00:39:11.480 align:middle line:84% or relative min happens at an endpoint. 00:39:11.480 --> 00:39:16.790 align:middle line:84% Think of f of x equals x on 0, 1. 00:39:16.790 --> 00:39:20.090 align:middle line:84% Then it has a absolute minimum at 0, absolute maximum at 1, 00:39:20.090 --> 00:39:23.060 align:middle line:84% and the derivative is 1 at both points, not 0. 00:39:23.060 --> 00:39:25.880 align:middle line:84% So this is only for relative min, 00:39:25.880 --> 00:39:28.670 align:middle line:84% relative maxes that occur inside-- 00:39:28.670 --> 00:39:32.750 align:middle line:84% strictly inside the interval that the function is defined on 00:39:32.750 --> 00:39:35.400 align:middle line:90% and where it's differentiable. 00:39:35.400 --> 00:39:38.580 align:middle line:90% 00:39:38.580 --> 00:39:49.310 align:middle line:84% So now, we have Rolle's theorem, which is essentially 00:39:49.310 --> 00:39:53.720 align:middle line:90% the mean value theorem rotated. 00:39:53.720 --> 00:39:55.970 align:middle line:84% We'll get the mean value theorem from Rolle's theorem. 00:39:55.970 --> 00:40:07.500 align:middle line:84% And which states the following, let f be from a, b to R, 00:40:07.500 --> 00:40:08.700 align:middle line:90% be continuous. 00:40:08.700 --> 00:40:14.400 align:middle line:90% 00:40:14.400 --> 00:40:23.430 align:middle line:84% So it's continuous at every point in a, b, differentiable 00:40:23.430 --> 00:40:27.787 align:middle line:90% at every point inside of a, b. 00:40:27.787 --> 00:40:29.620 align:middle line:84% It could be differentiable at the endpoints. 00:40:29.620 --> 00:40:30.120 align:middle line:90% That's fine. 00:40:30.120 --> 00:40:33.360 align:middle line:84% But it has to be differentiable on the open interval a, b. 00:40:33.360 --> 00:40:38.585 align:middle line:90% 00:40:38.585 --> 00:40:43.710 align:middle line:84% If f of b, a equals f of b equals 0, 00:40:43.710 --> 00:40:51.225 align:middle line:84% then there exists a point c in the open interval a, b-- 00:40:51.225 --> 00:40:53.950 align:middle line:90% and this is important-- 00:40:53.950 --> 00:40:58.570 align:middle line:90% such that f prime of c equals 0. 00:40:58.570 --> 00:41:02.200 align:middle line:84% And what's the what's the picture that goes with this? 00:41:02.200 --> 00:41:03.670 align:middle line:90% I'm sure you've seen it before. 00:41:03.670 --> 00:41:08.030 align:middle line:90% 00:41:08.030 --> 00:41:16.750 align:middle line:84% We draw a function that looks kind of like sine or cosine. 00:41:16.750 --> 00:41:20.530 align:middle line:84% But anyways, function which is 0 here, 0 here. 00:41:20.530 --> 00:41:27.550 align:middle line:84% Then there has to exist a point where the tangent is vertical. 00:41:27.550 --> 00:41:31.925 align:middle line:84% In fact, here it can occur at two different points. 00:41:31.925 --> 00:41:34.355 align:middle line:90% 00:41:34.355 --> 00:41:35.730 align:middle line:84% But something that's already kind 00:41:35.730 --> 00:41:38.160 align:middle line:84% of giving the game away as far as how we'll prove this 00:41:38.160 --> 00:41:44.190 align:middle line:84% is, let's look at where the function takes 00:41:44.190 --> 00:41:46.170 align:middle line:90% a maximum and a minimum. 00:41:46.170 --> 00:41:50.000 align:middle line:90% 00:41:50.000 --> 00:41:51.855 align:middle line:90% That is the heart of the proof. 00:41:51.855 --> 00:41:56.100 align:middle line:90% 00:41:56.100 --> 00:41:58.050 align:middle line:84% And why can we-- why does it even 00:41:58.050 --> 00:42:00.180 align:middle line:84% have an absolute maximum or absolute minimum? 00:42:00.180 --> 00:42:03.060 align:middle line:90% 00:42:03.060 --> 00:42:05.030 align:middle line:84% So I guess I didn't say this over there 00:42:05.030 --> 00:42:07.070 align:middle line:84% when I discussed relative min and relative max. 00:42:07.070 --> 00:42:11.790 align:middle line:84% But an absolute maximum is also a relative maximum. 00:42:11.790 --> 00:42:15.305 align:middle line:84% And absolute minimum is also a relative minimum. 00:42:15.305 --> 00:42:18.290 align:middle line:90% 00:42:18.290 --> 00:42:23.230 align:middle line:84% So let me make this-- this is kind of a late remark. 00:42:23.230 --> 00:42:30.340 align:middle line:90% 00:42:30.340 --> 00:42:48.260 align:middle line:84% Absolute max is a relative max, absolute mins 00:42:48.260 --> 00:42:49.310 align:middle line:90% is a relative min. 00:42:49.310 --> 00:42:54.120 align:middle line:90% 00:42:54.120 --> 00:42:58.420 align:middle line:84% So this picture is kind of given the game away already. 00:42:58.420 --> 00:43:02.020 align:middle line:84% Let's look at where the function takes a max and a min. 00:43:02.020 --> 00:43:06.030 align:middle line:84% And why does f take a max or a min? 00:43:06.030 --> 00:43:08.430 align:middle line:84% It's because f is continuous on this closed and bounded 00:43:08.430 --> 00:43:08.930 align:middle line:90% interval. 00:43:08.930 --> 00:43:13.050 align:middle line:84% This is something we proved when we were discussing continuity, 00:43:13.050 --> 00:43:14.860 align:middle line:90% the min max theorem. 00:43:14.860 --> 00:43:18.510 align:middle line:90% So let's give the proof. 00:43:18.510 --> 00:43:27.150 align:middle line:90% 00:43:27.150 --> 00:43:42.260 align:middle line:84% Since f is continuous on a, b, f achieves a relative max 00:43:42.260 --> 00:43:56.940 align:middle line:84% at some point c1 in a, b, and relative min at c2 in a, b. 00:43:56.940 --> 00:44:05.080 align:middle line:90% 00:44:05.080 --> 00:44:09.705 align:middle line:90% If there exists an x-- 00:44:09.705 --> 00:44:10.205 align:middle line:90% also. 00:44:10.205 --> 00:44:15.700 align:middle line:90% 00:44:15.700 --> 00:44:21.210 align:middle line:84% If of f of c1 is positive, then what can you say? 00:44:21.210 --> 00:44:24.290 align:middle line:84% Well, c1 cannot be one of the endpoints, 00:44:24.290 --> 00:44:29.240 align:middle line:84% because at the endpoints, f of a equals f of b equals 0. 00:44:29.240 --> 00:44:32.000 align:middle line:90% 00:44:32.000 --> 00:44:35.630 align:middle line:84% This implies c1 is in the open interval 00:44:35.630 --> 00:44:39.950 align:middle line:84% a, b, which implies by the previous theorem, f prime 00:44:39.950 --> 00:44:42.890 align:middle line:90% of c1 equals 0. 00:44:42.890 --> 00:44:47.250 align:middle line:90% So here you-- c equals c1. 00:44:47.250 --> 00:44:51.990 align:middle line:90% 00:44:51.990 --> 00:44:58.500 align:middle line:84% If I'm at the minimum now, and this is less than 0, 00:44:58.500 --> 00:45:07.370 align:middle line:84% then again, this implies that c2 is in the open interval a, 00:45:07.370 --> 00:45:15.300 align:middle line:84% b, which implies since f has a minimum at c2, f prime of c2 00:45:15.300 --> 00:45:17.310 align:middle line:84% equals 0 by the previous theorem. 00:45:17.310 --> 00:45:19.380 align:middle line:84% And therefore, we could take c equals c2. 00:45:19.380 --> 00:45:24.320 align:middle line:90% 00:45:24.320 --> 00:45:29.930 align:middle line:84% If-- now, this is two cases-- f of c1, if either f of c1 00:45:29.930 --> 00:45:32.600 align:middle line:84% is positive or f of c2 is less than 0, 00:45:32.600 --> 00:45:42.920 align:middle line:84% then we have the result. If f of c1 is less than or equal to 0, 00:45:42.920 --> 00:45:46.040 align:middle line:84% is less than or equal to f of c2, 00:45:46.040 --> 00:45:50.270 align:middle line:84% now remember, f achieves a max that c1. 00:45:50.270 --> 00:45:52.253 align:middle line:90% F achieves a min at c2. 00:45:52.253 --> 00:45:54.170 align:middle line:84% And therefore, this should always be less than 00:45:54.170 --> 00:45:54.980 align:middle line:90% or equal to this. 00:45:54.980 --> 00:46:08.110 align:middle line:90% 00:46:08.110 --> 00:46:10.615 align:middle line:90% If the max and the min-- 00:46:10.615 --> 00:46:16.620 align:middle line:90% 00:46:16.620 --> 00:46:23.920 align:middle line:84% so first off, I want to make sure I'm 00:46:23.920 --> 00:46:25.300 align:middle line:90% getting my logic correct here. 00:46:25.300 --> 00:46:44.980 align:middle line:90% 00:46:44.980 --> 00:46:52.285 align:middle line:84% So rather than state it this way, let's do it like this. 00:46:52.285 --> 00:47:02.850 align:middle line:90% 00:47:02.850 --> 00:47:05.300 align:middle line:84% So in terms of the picture here, this 00:47:05.300 --> 00:47:12.440 align:middle line:84% would be c1, c2, where we achieve a max and a min. 00:47:12.440 --> 00:47:36.060 align:middle line:84% Then in the final case that f of c1 and then, so 00:47:36.060 --> 00:47:37.530 align:middle line:90% this is the max. 00:47:37.530 --> 00:47:41.385 align:middle line:84% 0 is equal to f evaluated at a point. 00:47:41.385 --> 00:47:52.130 align:middle line:90% 00:47:52.130 --> 00:48:08.960 align:middle line:90% Then-- all right. 00:48:08.960 --> 00:48:12.990 align:middle line:84% So let me go back to what I was going to say a minute ago. 00:48:12.990 --> 00:48:18.050 align:middle line:84% So in the last case, that f of c1 is less than or equal to 0 00:48:18.050 --> 00:48:19.880 align:middle line:90% is less than f of c2. 00:48:19.880 --> 00:48:27.530 align:middle line:84% Now, remember, this is the min, this is the max of f. 00:48:27.530 --> 00:48:32.900 align:middle line:84% So this should always sit above f of c2. 00:48:32.900 --> 00:48:40.490 align:middle line:84% So that implies that f of c1 equals f of c2. 00:48:40.490 --> 00:48:42.620 align:middle line:84% And therefore, the max and min equal each other. 00:48:42.620 --> 00:48:46.400 align:middle line:90% 00:48:46.400 --> 00:48:55.110 align:middle line:84% Which implies for all x and a, b, since f of c1 00:48:55.110 --> 00:49:01.048 align:middle line:84% is the minimum, or f of c2 is the-- 00:49:01.048 --> 00:49:09.350 align:middle line:90% 00:49:09.350 --> 00:49:12.810 align:middle line:84% so I started off with something here, there, and in between, 00:49:12.810 --> 00:49:22.460 align:middle line:84% which implies for all x in a, b, f of x equals f of c2. 00:49:22.460 --> 00:49:24.500 align:middle line:90% So that means f is constant. 00:49:24.500 --> 00:49:31.200 align:middle line:90% 00:49:31.200 --> 00:49:33.750 align:middle line:84% And we know what the derivative of a constant function is. 00:49:33.750 --> 00:49:35.820 align:middle line:90% It's just 0. 00:49:35.820 --> 00:49:36.570 align:middle line:90% So we could take-- 00:49:36.570 --> 00:49:41.415 align:middle line:90% 00:49:41.415 --> 00:49:42.165 align:middle line:90% say, the midpoint. 00:49:42.165 --> 00:49:48.440 align:middle line:90% 00:49:48.440 --> 00:49:50.700 align:middle line:90% And that's the end. 00:49:50.700 --> 00:49:54.690 align:middle line:84% Sorry that I kind of fumbled through that for a minute, 00:49:54.690 --> 00:49:57.370 align:middle line:90% but-- 00:49:57.370 --> 00:49:58.870 align:middle line:90% OK. 00:49:58.870 --> 00:50:03.370 align:middle line:84% So for any function which is 0 at the endpoints, 00:50:03.370 --> 00:50:09.130 align:middle line:84% there has to be a point in between where the function is-- 00:50:09.130 --> 00:50:12.040 align:middle line:84% the derivative of the function equals 0. 00:50:12.040 --> 00:50:19.460 align:middle line:84% Now, again, we should come back to when we see a theorem, 00:50:19.460 --> 00:50:21.640 align:middle line:84% we should pick it apart a little bit 00:50:21.640 --> 00:50:23.275 align:middle line:84% to see what's necessary, what's not. 00:50:23.275 --> 00:50:26.898 align:middle line:90% 00:50:26.898 --> 00:50:29.482 align:middle line:84% We had these two hypotheses coming in, 00:50:29.482 --> 00:50:31.690 align:middle line:84% that if the function is continuous and differentiable 00:50:31.690 --> 00:50:37.030 align:middle line:84% on this open interval a, b, are these necessary? 00:50:37.030 --> 00:50:59.740 align:middle line:84% So for example, let's say I have the function-- 00:50:59.740 --> 00:51:05.560 align:middle line:84% if I look at the function 1 minus the absolute value of x. 00:51:05.560 --> 00:51:17.360 align:middle line:84% So at minus 1, 1, 1, this is a continuous function 00:51:17.360 --> 00:51:25.305 align:middle line:90% on minus 1, 1. 00:51:25.305 --> 00:51:28.920 align:middle line:90% 00:51:28.920 --> 00:51:32.310 align:middle line:84% And f of minus 1 equals f of 1 equals 0. 00:51:32.310 --> 00:51:36.000 align:middle line:84% But there is no point where f prime of c equals 0. 00:51:36.000 --> 00:51:39.160 align:middle line:84% What hypothesis did I leave out for this function? 00:51:39.160 --> 00:51:40.320 align:middle line:90% The fact that it's-- 00:51:40.320 --> 00:51:42.930 align:middle line:84% the hypothesis that it's differentiable at every point 00:51:42.930 --> 00:51:44.370 align:middle line:90% in between. 00:51:44.370 --> 00:51:46.200 align:middle line:84% This function is not differentiable. 00:51:46.200 --> 00:51:58.280 align:middle line:90% 00:51:58.280 --> 00:52:01.340 align:middle line:84% But there's no point where the derivative equals 0. 00:52:01.340 --> 00:52:05.420 align:middle line:84% So this example tells you that the hypothesis 00:52:05.420 --> 00:52:10.550 align:middle line:84% that the function is differentiable 00:52:10.550 --> 00:52:14.750 align:middle line:84% on the interval minus 1, 1, or a, b, 00:52:14.750 --> 00:52:17.585 align:middle line:84% is necessary for the theorem to be true. 00:52:17.585 --> 00:52:18.710 align:middle line:90% That's what I'm getting at. 00:52:18.710 --> 00:52:26.590 align:middle line:90% 00:52:26.590 --> 00:52:30.610 align:middle line:84% Now, the other hypothesis that the function 00:52:30.610 --> 00:52:37.420 align:middle line:84% is continuous on the closed interval is also necessary. 00:52:37.420 --> 00:52:38.455 align:middle line:90% Let's say I look at-- 00:52:38.455 --> 00:52:43.470 align:middle line:90% 00:52:43.470 --> 00:52:46.510 align:middle line:84% and this is pretty easy to come up with a counter-example, 00:52:46.510 --> 00:52:47.640 align:middle line:90% if I don't assume that. 00:52:47.640 --> 00:52:54.180 align:middle line:90% 00:52:54.180 --> 00:53:06.820 align:middle line:84% Let's take the function which is-- 00:53:06.820 --> 00:53:22.480 align:middle line:90% 00:53:22.480 --> 00:53:25.810 align:middle line:84% the function that's x minus 1 for x not equal to minus 1, 00:53:25.810 --> 00:53:29.470 align:middle line:90% and 0 for x equal to minus 1. 00:53:29.470 --> 00:53:31.240 align:middle line:84% So does is this function look like? 00:53:31.240 --> 00:53:41.810 align:middle line:90% 00:53:41.810 --> 00:53:51.860 align:middle line:84% So this function here is differentiable on minus 1, 1, 00:53:51.860 --> 00:53:56.600 align:middle line:84% f of minus 1 equals f of 1 equals 0. 00:53:56.600 --> 00:54:04.250 align:middle line:84% But again, there's no point c in minus 1, 1 so 00:54:04.250 --> 00:54:08.660 align:middle line:90% that f prime of c equals 0. 00:54:08.660 --> 00:54:13.940 align:middle line:84% Because the derivative in between minus 1, 1 is just 1. 00:54:13.940 --> 00:54:17.780 align:middle line:84% So both of these hypotheses, that f 00:54:17.780 --> 00:54:20.690 align:middle line:84% is continuous on the closed interval and differentiable 00:54:20.690 --> 00:54:23.720 align:middle line:84% on the open interval, are necessary for this theorem 00:54:23.720 --> 00:54:24.590 align:middle line:90% to be true. 00:54:24.590 --> 00:54:27.827 align:middle line:84% If you drop either of those hypotheses, 00:54:27.827 --> 00:54:28.910 align:middle line:90% then the theorem is false. 00:54:28.910 --> 00:54:45.800 align:middle line:90% 00:54:45.800 --> 00:54:49.340 align:middle line:84% So let's rotate this picture here and arrive 00:54:49.340 --> 00:54:50.480 align:middle line:90% at the mean value theorem. 00:54:50.480 --> 00:55:03.120 align:middle line:90% 00:55:03.120 --> 00:55:09.040 align:middle line:84% Let f from a, b to R be continuous. 00:55:09.040 --> 00:55:16.600 align:middle line:84% So it's continuous on the closed interval a, b, differentiable 00:55:16.600 --> 00:55:17.100 align:middle line:90% on a, b. 00:55:17.100 --> 00:55:25.750 align:middle line:90% 00:55:25.750 --> 00:55:29.800 align:middle line:84% Then there exists a c in the open interval 00:55:29.800 --> 00:55:40.166 align:middle line:90% a, b such that f prime of c-- 00:55:40.166 --> 00:55:41.410 align:middle line:90% let me write it this way-- 00:55:41.410 --> 00:55:47.002 align:middle line:90% 00:55:47.002 --> 00:55:54.800 align:middle line:84% f of b minus f of a times b minus a. 00:55:54.800 --> 00:55:58.500 align:middle line:90% 00:55:58.500 --> 00:56:04.560 align:middle line:84% So why do I refer to it as a rotated picture and shifted 00:56:04.560 --> 00:56:05.820 align:middle line:90% version of Rolle's theorem? 00:56:05.820 --> 00:56:08.560 align:middle line:90% 00:56:08.560 --> 00:56:11.060 align:middle line:90% So there's a, b. 00:56:11.060 --> 00:56:14.450 align:middle line:90% Here's f of a, f of b. 00:56:14.450 --> 00:56:20.860 align:middle line:84% And let's say that's how the function looks. 00:56:20.860 --> 00:56:27.300 align:middle line:84% Now, the slope of the line connecting f of b to f of a 00:56:27.300 --> 00:56:29.602 align:middle line:90% is exactly f of b minus f of a. 00:56:29.602 --> 00:56:34.090 align:middle line:90% 00:56:34.090 --> 00:56:40.080 align:middle line:84% So the slope of that line is f of b minus f of a. 00:56:40.080 --> 00:56:45.980 align:middle line:84% And what we're stating is that there's a point c, 00:56:45.980 --> 00:56:53.050 align:middle line:84% so that the tangent at that point is parallel to that line, 00:56:53.050 --> 00:57:00.220 align:middle line:84% has the same slope as the line connecting f of b to f of a. 00:57:00.220 --> 00:57:09.480 align:middle line:90% 00:57:09.480 --> 00:57:13.800 align:middle line:84% So this theorem really does reduce to Rolle's theorem. 00:57:13.800 --> 00:57:23.020 align:middle line:84% So let me define a function g from a, b 00:57:23.020 --> 00:57:30.770 align:middle line:84% to R, which satisfies the hypotheses of Rolle's theorem 00:57:30.770 --> 00:57:35.050 align:middle line:84% and will give us what we want, basically. 00:57:35.050 --> 00:57:43.270 align:middle line:84% So g of x equals f of x minus f of b, 00:57:43.270 --> 00:57:54.290 align:middle line:84% plus f of b minus f of a over b minus a times b minus x. 00:57:54.290 --> 00:57:57.960 align:middle line:90% 00:57:57.960 --> 00:58:00.210 align:middle line:84% Basically, what this function g does is, 00:58:00.210 --> 00:58:04.470 align:middle line:84% it takes the function now and rotates it, 00:58:04.470 --> 00:58:07.500 align:middle line:90% and shifts it down, so that f-- 00:58:07.500 --> 00:58:10.215 align:middle line:84% these two points then coincide and give you 0. 00:58:10.215 --> 00:58:15.920 align:middle line:90% 00:58:15.920 --> 00:58:24.190 align:middle line:84% Then g is continuous on the closed interval a, b. 00:58:24.190 --> 00:58:24.690 align:middle line:90% Why? 00:58:24.690 --> 00:58:27.960 align:middle line:84% Because it's a sum of two continuous functions on a, b. 00:58:27.960 --> 00:58:30.570 align:middle line:90% f of x is continuous on a, b. 00:58:30.570 --> 00:58:32.220 align:middle line:90% And this is just a constant. 00:58:32.220 --> 00:58:33.580 align:middle line:90% That's just a constant. 00:58:33.580 --> 00:58:35.580 align:middle line:90% This is a polynomial. 00:58:35.580 --> 00:58:47.500 align:middle line:84% So g is continuous on a, b, and differentiable on a, b, 00:58:47.500 --> 00:58:49.240 align:middle line:90% again, because f is. 00:58:49.240 --> 00:58:52.630 align:middle line:84% F is differentiable on the open interval a, b. 00:58:52.630 --> 00:58:55.240 align:middle line:84% This part is differentiable everywhere 00:58:55.240 --> 00:58:59.350 align:middle line:84% so the sum of two differentiable functions is differential. 00:58:59.350 --> 00:59:01.030 align:middle line:84% So g of x is differentiable on a, b. 00:59:01.030 --> 00:59:04.580 align:middle line:90% 00:59:04.580 --> 00:59:09.010 align:middle line:84% So continuous on a, b and differentiable on a, b. 00:59:09.010 --> 00:59:12.902 align:middle line:90% 00:59:12.902 --> 00:59:15.110 align:middle line:84% And since I said, we're going to use Rolle's theorem, 00:59:15.110 --> 00:59:16.767 align:middle line:90% let's compute g of a. 00:59:16.767 --> 00:59:22.280 align:middle line:84% This is equal to f of a minus f of b, 00:59:22.280 --> 00:59:30.800 align:middle line:84% plus f of b minus f of a, over b minus a, times b minus a. 00:59:30.800 --> 00:59:32.150 align:middle line:90% That cancels with that. 00:59:32.150 --> 00:59:34.770 align:middle line:84% Minus f of b cancels with f of b. 00:59:34.770 --> 00:59:37.790 align:middle line:84% f of a cancels with minus f of a, and we get 0. 00:59:37.790 --> 00:59:41.391 align:middle line:90% 00:59:41.391 --> 00:59:45.080 align:middle line:84% And then, g of b, this is even easier to see. 00:59:45.080 --> 00:59:46.580 align:middle line:84% If I just stick this in, this is f 00:59:46.580 --> 00:59:56.690 align:middle line:84% of b minus f of b plus f of b minus f of a, over b minus a, 00:59:56.690 --> 00:59:58.460 align:middle line:90% b minus b. 00:59:58.460 --> 01:00:00.290 align:middle line:90% And this also equals 0. 01:00:00.290 --> 01:00:02.930 align:middle line:90% So g of a and g of b are 0. 01:00:02.930 --> 01:00:05.180 align:middle line:84% The function is continuous on the closed interval. 01:00:05.180 --> 01:00:07.160 align:middle line:84% It's differentiable on the open interval. 01:00:07.160 --> 01:00:15.073 align:middle line:84% And therefore, by Rolle's theorem-- 01:00:15.073 --> 01:00:17.490 align:middle line:84% I think there's supposed to be some sort of accent over O, 01:00:17.490 --> 01:00:22.360 align:middle line:84% but I can't remember, and I've forgotten it already. 01:00:22.360 --> 01:00:28.600 align:middle line:84% By Rolle's theorem, there exist a point c in a, b, such 01:00:28.600 --> 01:00:40.850 align:middle line:84% that 0 is equal to g prime of c, which is equal to-- 01:00:40.850 --> 01:00:43.430 align:middle line:84% we take g and actually compute what the derivative is 01:00:43.430 --> 01:00:49.010 align:middle line:90% at c, is equal to f prime of c. 01:00:49.010 --> 01:00:51.650 align:middle line:84% That's a constant derivative of that is 0. 01:00:51.650 --> 01:00:55.310 align:middle line:84% This times this, the derivative of x evaluated at c just 01:00:55.310 --> 01:01:05.590 align:middle line:84% gives me minus f of b minus f of a, times b minus a. 01:01:05.590 --> 01:01:08.700 align:middle line:90% And that's it. 01:01:08.700 --> 01:01:12.330 align:middle line:84% 0 is equal to f prime of c minus what we want. 01:01:12.330 --> 01:01:15.485 align:middle line:84% We move that to the other side, multiply through by b minus a, 01:01:15.485 --> 01:01:16.110 align:middle line:90% and we're done. 01:01:16.110 --> 01:01:34.790 align:middle line:90% 01:01:34.790 --> 01:01:41.480 align:middle line:84% So some first very nice applications 01:01:41.480 --> 01:01:43.160 align:middle line:84% of the mean value theorem, which I 01:01:43.160 --> 01:01:46.280 align:middle line:84% think you learned from calculus are the following. 01:01:46.280 --> 01:02:13.080 align:middle line:90% 01:02:13.080 --> 01:02:18.080 align:middle line:84% Let f from I to R be differentiable, 01:02:18.080 --> 01:02:24.010 align:middle line:84% meaning it's differentiable at every point in the interval I. 01:02:24.010 --> 01:02:27.130 align:middle line:90% Then we have two conclusions. 01:02:27.130 --> 01:02:36.860 align:middle line:84% f is increasing, meaning we recall what that means, 01:02:36.860 --> 01:02:42.860 align:middle line:84% for f to be increasing, that means that if x is less than y, 01:02:42.860 --> 01:02:47.150 align:middle line:84% this implies f of x is less than or equal to f of y. 01:02:47.150 --> 01:02:50.960 align:middle line:90% 01:02:50.960 --> 01:03:01.160 align:middle line:84% This is equivalent to for all x in I, 01:03:01.160 --> 01:03:03.710 align:middle line:84% f prime of x is bigger than or equal to 0. 01:03:03.710 --> 01:03:06.890 align:middle line:84% So remember, f prime is the rate of change. 01:03:06.890 --> 01:03:08.850 align:middle line:84% If the rate of change is always non-negative, 01:03:08.850 --> 01:03:10.590 align:middle line:90% then f has to be growing. 01:03:10.590 --> 01:03:19.970 align:middle line:84% But this says, f is decreasing, meaning x less than y 01:03:19.970 --> 01:03:25.760 align:middle line:84% implies f of x is bigger than or equal to f of y. 01:03:25.760 --> 01:03:30.080 align:middle line:84% This is equivalent to since a function is increasing 01:03:30.080 --> 01:03:32.780 align:middle line:84% if and only if minus f is decreasing, 01:03:32.780 --> 01:03:34.250 align:middle line:90% this inequality should flip. 01:03:34.250 --> 01:03:52.570 align:middle line:90% 01:03:52.570 --> 01:03:56.950 align:middle line:84% So I'm going to do increasing-- decreasing follows from number 01:03:56.950 --> 01:03:58.870 align:middle line:90% one by taking minus f. 01:03:58.870 --> 01:04:09.130 align:middle line:90% 01:04:09.130 --> 01:04:13.040 align:middle line:90% So how much time do I have? 01:04:13.040 --> 01:04:13.810 align:middle line:90% I still have time. 01:04:13.810 --> 01:04:21.030 align:middle line:90% 01:04:21.030 --> 01:04:28.012 align:middle line:84% So suppose first that f prime of x is always non-negative. 01:04:28.012 --> 01:04:32.120 align:middle line:90% 01:04:32.120 --> 01:04:43.550 align:middle line:84% Let a, b be in I. [SNEEZES] Excuse me. 01:04:43.550 --> 01:04:47.872 align:middle line:90% 01:04:47.872 --> 01:04:50.330 align:middle line:84% We now want to show f of a is less than or equal to f of b. 01:04:50.330 --> 01:04:55.390 align:middle line:90% 01:04:55.390 --> 01:05:01.560 align:middle line:84% Then f is continuous on this smaller interval a, b. 01:05:01.560 --> 01:05:02.400 align:middle line:90% Why? 01:05:02.400 --> 01:05:04.950 align:middle line:84% Because f is differentiable at every point in I. 01:05:04.950 --> 01:05:11.070 align:middle line:84% And therefore, it is continuous at every point in I. 01:05:11.070 --> 01:05:24.250 align:middle line:84% So in particular on this smaller interval, and differentiable 01:05:24.250 --> 01:05:33.190 align:middle line:84% on a, b, which implies there exists a c in a, b-- 01:05:33.190 --> 01:05:42.370 align:middle line:84% So by the mean value theorem, such that f 01:05:42.370 --> 01:05:49.630 align:middle line:84% of b minus f of a equals f prime of c times b minus a. 01:05:49.630 --> 01:05:53.410 align:middle line:84% And now, since the derivative is always non-negative 01:05:53.410 --> 01:05:57.400 align:middle line:84% and b minus a is positive, this is bigger than or equal to 0. 01:05:57.400 --> 01:05:59.020 align:middle line:90% I.e. 01:05:59.020 --> 01:06:01.825 align:middle line:84% if a b minus f of a is bigger than or equal to 0. 01:06:01.825 --> 01:06:07.280 align:middle line:90% 01:06:07.280 --> 01:06:08.840 align:middle line:90% So we prove in one direction. 01:06:08.840 --> 01:06:19.790 align:middle line:84% We've proven that if you like that the function's 01:06:19.790 --> 01:06:22.400 align:middle line:84% derivative being non-negative implies that the function is 01:06:22.400 --> 01:06:24.910 align:middle line:90% increasing. 01:06:24.910 --> 01:06:26.760 align:middle line:84% So now let's prove the opposite direction. 01:06:26.760 --> 01:06:35.010 align:middle line:90% 01:06:35.010 --> 01:06:54.100 align:middle line:84% So suppose f is increasing and c is in I. 01:06:54.100 --> 01:07:07.100 align:middle line:84% Let xn be a sequence in I, such that xn converges to c. 01:07:07.100 --> 01:07:09.695 align:middle line:90% And one of two things holds. 01:07:09.695 --> 01:07:12.770 align:middle line:90% 01:07:12.770 --> 01:07:28.580 align:middle line:84% And either let's say, a, all n xn is less than c, 01:07:28.580 --> 01:07:32.900 align:middle line:90% or b, all n xn is bigger than c. 01:07:32.900 --> 01:07:35.480 align:middle line:84% And we can always find such a sequence given a point 01:07:35.480 --> 01:07:39.920 align:middle line:90% C in the interval I. 01:07:39.920 --> 01:07:48.520 align:middle line:84% So if I, so let's say, it's a closed interval. 01:07:48.520 --> 01:07:51.280 align:middle line:90% And c is in here. 01:07:51.280 --> 01:07:55.780 align:middle line:84% Then we can find a sequence xn converging to c, 01:07:55.780 --> 01:07:57.505 align:middle line:90% approaching c from the left. 01:07:57.505 --> 01:08:00.430 align:middle line:90% 01:08:00.430 --> 01:08:03.010 align:middle line:84% If c is one of the endpoints, then we 01:08:03.010 --> 01:08:06.520 align:middle line:84% can find a sequence from I approaching c from the right. 01:08:06.520 --> 01:08:07.900 align:middle line:90% That's b. 01:08:07.900 --> 01:08:09.790 align:middle line:84% And the same thing if c is over here we 01:08:09.790 --> 01:08:13.330 align:middle line:84% can find a sequence approaching c, strictly from the left. 01:08:13.330 --> 01:08:15.940 align:middle line:90% 01:08:15.940 --> 01:08:18.400 align:middle line:84% So there always exists such a sequence. 01:08:18.400 --> 01:08:28.960 align:middle line:90% 01:08:28.960 --> 01:08:29.700 align:middle line:90% Always exists. 01:08:29.700 --> 01:08:32.979 align:middle line:90% 01:08:32.979 --> 01:08:38.795 align:middle line:84% And this is because, I is an interval. 01:08:38.795 --> 01:08:53.500 align:middle line:90% 01:08:53.500 --> 01:09:00.750 align:middle line:84% And if the sequence satisfies case a, and we look at f prime, 01:09:00.750 --> 01:09:08.569 align:middle line:84% then we get that f of xn is less than 01:09:08.569 --> 01:09:14.120 align:middle line:90% or equal to f of c for all n. 01:09:14.120 --> 01:09:17.090 align:middle line:84% And let me in fact write this a little bit differently. 01:09:17.090 --> 01:09:19.850 align:middle line:84% Since-- so here we're using that f as increasing. 01:09:19.850 --> 01:09:28.880 align:middle line:90% 01:09:28.880 --> 01:09:34.689 align:middle line:84% Since f is increasing, which implies 01:09:34.689 --> 01:09:42.700 align:middle line:84% for all n, f of xn minus f of c over x of n minus c-- 01:09:42.700 --> 01:09:48.729 align:middle line:84% now, this thing on top is negative. xn is less than c. 01:09:48.729 --> 01:09:53.830 align:middle line:84% So this thing on the bottom is also negative. 01:09:53.830 --> 01:09:57.940 align:middle line:84% And therefore, this thing is, positive-- 01:09:57.940 --> 01:10:00.860 align:middle line:90% non-negative, I mean. 01:10:00.860 --> 01:10:03.020 align:middle line:84% And therefore, the limit must also be negative. 01:10:03.020 --> 01:10:05.610 align:middle line:90% 01:10:05.610 --> 01:10:08.280 align:middle line:84% But that limit is just a derivative evaluated at c. 01:10:08.280 --> 01:10:21.070 align:middle line:90% 01:10:21.070 --> 01:10:23.230 align:middle line:90% And case b is kind of similar. 01:10:23.230 --> 01:10:27.040 align:middle line:90% I mean, not kind of, it is. 01:10:27.040 --> 01:10:32.500 align:middle line:84% In case b, we get that for all n, f of xn, 01:10:32.500 --> 01:10:34.600 align:middle line:84% now, since x tends to the right of c-- 01:10:34.600 --> 01:10:37.890 align:middle line:90% 01:10:37.890 --> 01:10:39.140 align:middle line:90% let me write it this way. 01:10:39.140 --> 01:10:42.110 align:middle line:90% 01:10:42.110 --> 01:10:46.040 align:middle line:84% This minus f of c is bigger than or equal to 0, 01:10:46.040 --> 01:10:51.950 align:middle line:84% which implies that f prime of c, which 01:10:51.950 --> 01:10:56.360 align:middle line:84% is the limit as n goes to infinity of f of xn 01:10:56.360 --> 01:11:01.040 align:middle line:90% minus f of c over xn minus c. 01:11:01.040 --> 01:11:02.798 align:middle line:90% What's on top is non-negative. 01:11:02.798 --> 01:11:04.340 align:middle line:84% And what's on bottom is non-negative, 01:11:04.340 --> 01:11:08.120 align:middle line:90% because xn is bigger than c. 01:11:08.120 --> 01:11:12.382 align:middle line:84% So bigger than 0, bigger than 0, bigger than 0. 01:11:12.382 --> 01:11:19.660 align:middle line:90% 01:11:19.660 --> 01:11:22.210 align:middle line:84% Thus in either case, we get that f 01:11:22.210 --> 01:11:24.533 align:middle line:84% prime of c is bigger than or equal to 0. 01:11:24.533 --> 01:11:28.915 align:middle line:90% 01:11:28.915 --> 01:11:31.290 align:middle line:84% And therefore, we've proven that the derivative is always 01:11:31.290 --> 01:11:33.540 align:middle line:90% non-negative. 01:11:33.540 --> 01:11:36.570 align:middle line:90% So that's one. 01:11:36.570 --> 01:11:48.870 align:middle line:84% And two, f is decreasing if and only 01:11:48.870 --> 01:11:56.910 align:middle line:84% if the function minus f is increasing, which by what 01:11:56.910 --> 01:12:02.610 align:middle line:84% we've done from part one is if and only if minus f prime of x 01:12:02.610 --> 01:12:07.530 align:middle line:84% is bigger than or equal to 0 for all x in I. And then, 01:12:07.530 --> 01:12:15.750 align:middle line:84% multiplying through by this minus 1 gives me number two. 01:12:15.750 --> 01:12:22.570 align:middle line:90% 01:12:22.570 --> 01:12:26.620 align:middle line:90% And so, let me make one last-- 01:12:26.620 --> 01:12:28.960 align:middle line:84% I know it's taboo to write on the back part. 01:12:28.960 --> 01:12:30.610 align:middle line:90% But I'm going to do it anyways. 01:12:30.610 --> 01:12:33.325 align:middle line:84% We have the very simple theorem which now follows. 01:12:33.325 --> 01:12:38.830 align:middle line:90% 01:12:38.830 --> 01:12:47.890 align:middle line:84% Let f from I to R be differentiable, 01:12:47.890 --> 01:12:53.630 align:middle line:90% then f is constant. 01:12:53.630 --> 01:12:58.690 align:middle line:84% So f of x equals f of y for all x and y in I, 01:12:58.690 --> 01:13:03.770 align:middle line:84% if and only if the derivative is identically 0. 01:13:03.770 --> 01:13:09.870 align:middle line:90% 01:13:09.870 --> 01:13:25.630 align:middle line:84% So f is constant if and only if f is both 01:13:25.630 --> 01:13:29.690 align:middle line:90% increasing and decreasing. 01:13:29.690 --> 01:13:37.030 align:middle line:90% 01:13:37.030 --> 01:13:40.600 align:middle line:84% Because it satisfies the equality sign in both of those. 01:13:40.600 --> 01:13:46.960 align:middle line:84% And by what we've done now, this means for all x in I, 01:13:46.960 --> 01:13:52.480 align:middle line:84% if prime of x is bigger than or equal to 0, and-- 01:13:52.480 --> 01:13:54.520 align:middle line:84% that's for the increasing part-- f prime 01:13:54.520 --> 01:13:55.990 align:middle line:90% of x is less than or equal to 0. 01:13:55.990 --> 01:13:58.610 align:middle line:90% 01:13:58.610 --> 01:14:09.530 align:middle line:84% And therefore, this is equivalent to f 01:14:09.530 --> 01:14:10.760 align:middle line:90% prime of x equals 0. 01:14:10.760 --> 01:14:17.820 align:middle line:90% 01:14:17.820 --> 01:14:20.030 align:middle line:90% So we'll stop there. 01:14:20.030 --> 01:14:27.000 align:middle line:90%