WEBVTT
00:00:01.320 --> 00:00:06.710
In this video, we’ll learn how to interpret and use the mean value theorem and Rolle’s theorem.
00:00:07.390 --> 00:00:09.780
We’ll begin by looking at how Rolle’s theorem works.
00:00:10.000 --> 00:00:16.990
And how it leads us into the mean value theorem, before considering a number of examples of the application of this theorem.
00:00:17.720 --> 00:00:22.800
Rolle’s theorem was first published in 1691 by French mathematician Michel Rolle.
00:00:23.140 --> 00:00:31.200
He was a vocal critic of calculus, calling it inaccurate and a collection of ingenious fallacies, though he did eventually change his opinion.
00:00:31.810 --> 00:00:37.610
Rolle’s theorem says that if 𝑓 is a function that satisfies the following three hypotheses.
00:00:37.920 --> 00:00:43.370
That is, it’s continuous on the closed interval 𝑎 to 𝑏 and differentiable on the open interval 𝑎 to 𝑏.
00:00:43.620 --> 00:00:45.940
And 𝑓 of 𝑎 is equal to 𝑓 of 𝑏.
00:00:46.330 --> 00:00:55.890
Then, there exists a number 𝑐 in the open interval 𝑎 to 𝑏, such that the derivative of 𝑓 evaluated at 𝑐, that’s 𝑓 prime of 𝑐, is equal to zero.
00:00:56.480 --> 00:01:04.680
In other words, if the function satisfies these criteria, then there’s a point on the graph in this closed interval for which the slope of the curve is zero.
00:01:04.680 --> 00:01:07.430
The tangent of the curve at this point is horizontal.
00:01:07.830 --> 00:01:09.710
It also has a clear physical meaning.
00:01:09.920 --> 00:01:14.940
If a body moves along a straight line and after a certain period it returns to the starting point.
00:01:15.240 --> 00:01:20.850
Then there’s a moment in this period of time where the instantaneous velocity of the body must be equal to zero.
00:01:21.530 --> 00:01:26.960
Now, this theorem is seldom used as it tells us of the existence of a solution, but not how to get there.
00:01:27.240 --> 00:01:31.590
It is, however, extremely useful in helping us to derive the mean value theorem.
00:01:31.850 --> 00:01:32.950
Let’s have a look at that.
00:01:33.460 --> 00:01:39.020
This theorem was first stated by another French mathematician, Joseph-Louis Lagrange.
00:01:39.420 --> 00:01:43.140
Though we don’t know whether he was as critical of calculus as Michel Rolle.
00:01:43.520 --> 00:01:52.450
It says that if 𝑓 of 𝑥 is a function which is continuous on some closed interval 𝑎 to 𝑏 and differentiable at every point of some open interval 𝑎 to 𝑏.
00:01:52.650 --> 00:01:55.710
Then there is a point 𝑐 in this open interval.
00:01:55.890 --> 00:02:04.430
Such that the derivative of 𝑓 evaluated at 𝑐, that’s 𝑓 prime of 𝑐, is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
00:02:04.770 --> 00:02:08.180
Notice that this right-hand side is essentially the slope formula.
00:02:08.440 --> 00:02:14.140
𝑚 equals 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one, written using function notation.
00:02:14.510 --> 00:02:27.400
And what this formula therefore tells us is that there’s some 𝑥-value in the open interval 𝑎 to 𝑏, where the slope of the tangent line is the same as the slope of the secant line passing through the two end points of the closed interval.
00:02:28.000 --> 00:02:29.110
Let’s prove this.
00:02:29.510 --> 00:02:38.070
We’ll let 𝑔 of 𝑥 be the secant line to 𝑓 of 𝑥 passing through the two end points of our closed interval at 𝑎, 𝑓 of 𝑎 and 𝑏, 𝑓 of 𝑏.
00:02:38.420 --> 00:02:43.240
We can look to find the equation for 𝑔 of 𝑥 using the formula for the equation of a straight line.
00:02:43.240 --> 00:02:47.020
That’s 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one.
00:02:47.400 --> 00:02:49.760
We can let 𝑥 one be equal to 𝑎.
00:02:49.760 --> 00:02:51.970
And thus 𝑦 one is equal to 𝑓 of 𝑎.
00:02:52.250 --> 00:02:55.190
We can also change 𝑦 to 𝑔 of 𝑥.
00:02:55.460 --> 00:02:59.580
And we know that the slope 𝑚 is given by the change in 𝑦 over the change in 𝑥.
00:02:59.760 --> 00:03:02.880
That’s 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
00:03:03.280 --> 00:03:11.360
And we obtain that 𝑔 of 𝑥 minus 𝑓 of 𝑎 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎 times 𝑥 minus 𝑎.
00:03:11.830 --> 00:03:16.000
And we can solve for 𝑔 of 𝑥 by adding 𝑓 of 𝑎 to both sides.
00:03:16.390 --> 00:03:18.860
Next, we introduce a new function ℎ.
00:03:19.040 --> 00:03:25.260
And this is defined as the vertical distance between 𝑓 of 𝑥 and the secant line 𝑔 of 𝑥.
00:03:25.450 --> 00:03:29.850
So ℎ of 𝑥 is defined as 𝑓 of 𝑥 minus 𝑔 of 𝑥.
00:03:30.210 --> 00:03:35.240
Our next job is to substitute the expression for 𝑔 of 𝑥 into this equation for ℎ of 𝑥.
00:03:35.450 --> 00:03:37.190
But how is this useful?
00:03:37.600 --> 00:03:42.110
Well, notice that ℎ of 𝑎 and ℎ of 𝑏 are equal.
00:03:42.140 --> 00:03:49.620
They’re both zero since the vertical distance between the function 𝑓 of 𝑥 and the secant line is zero at these end points.
00:03:49.970 --> 00:03:52.350
And now we should be reminded of Rolle’s theorem.
00:03:52.870 --> 00:03:58.260
ℎ of 𝑥 is continuous and differentiable on the open interval 𝑎 to 𝑏.
00:03:58.730 --> 00:04:08.210
And so there must exist some value of 𝑥, 𝑐 in this interval, such that ℎ prime of 𝑐, the derivative of ℎ evaluated at 𝑐, is equal to zero.
00:04:08.540 --> 00:04:11.650
So let’s differentiate both sides of this equation.
00:04:12.010 --> 00:04:14.720
The derivative of ℎ is ℎ prime.
00:04:15.100 --> 00:04:17.620
And the derivative of 𝑓 is 𝑓 prime.
00:04:18.040 --> 00:04:20.610
Now let’s look carefully at everything inside this bracket.
00:04:20.860 --> 00:04:29.500
When we distribute these parentheses, we end up with this quotient, which is a constant being multiplied by 𝑥 and then being multiplied by another constant.
00:04:29.730 --> 00:04:32.580
Similarly, 𝑓 of 𝑎 is also a constant.
00:04:32.820 --> 00:04:35.430
We know that the derivative of a constant is zero.
00:04:35.830 --> 00:04:41.910
And we also know that the derivative of some constant multiplied by 𝑥 is simply that constant.
00:04:42.120 --> 00:04:48.840
So we see that ℎ prime of 𝑥 is equal to 𝑓 prime of 𝑥 minus 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
00:04:49.300 --> 00:04:58.510
From Rolle’s theorem, we can say that ℎ prime of 𝑐 must be equal to 𝑓 prime of 𝑐 minus 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎, which must be equal to zero.
00:04:59.010 --> 00:05:01.370
And we’ve actually proved the mean value theorem.
00:05:02.140 --> 00:05:10.870
If we solve for 𝑓 prime of 𝑐, we obtain that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
00:05:11.890 --> 00:05:18.610
Now that we’ve proved the mean value theorem, let’s have a look at an example of how to establish where the theorem actually applies.
00:05:19.470 --> 00:05:27.990
Does the mean value theorem apply for the function 𝑦 equals two 𝑥 cubed minus four 𝑥 plus seven over the closed interval zero to five?
00:05:28.670 --> 00:05:34.290
To use the mean value theorem, two things must be true about our function 𝑓 of 𝑥.
00:05:34.420 --> 00:05:37.820
It must be continuous over the closed interval 𝑎 to 𝑏.
00:05:38.230 --> 00:05:42.140
And it must be differentiable over the open interval 𝑎 to 𝑏.
00:05:42.750 --> 00:05:49.750
Well, the function two 𝑥 cubed minus four 𝑥 plus seven is indeed continuous over the closed interval zero to five.
00:05:50.330 --> 00:05:54.870
It’s a simple cubic graph that looks a little like this over our closed interval.
00:05:55.270 --> 00:06:00.480
And to check for the second condition, we’ll see what happens when we do differentiate with respect to 𝑥.
00:06:00.800 --> 00:06:05.330
The derivative of two 𝑥 cubed is three times two 𝑥 squared.
00:06:05.330 --> 00:06:06.330
That’s six 𝑥 squared.
00:06:06.620 --> 00:06:09.800
And the derivative of negative four 𝑥 is negative four.
00:06:10.040 --> 00:06:14.200
So we obtain d𝑦 by d𝑥 to be equal to six 𝑥 squared minus four.
00:06:14.600 --> 00:06:18.050
This is indeed defined over the open interval zero to five.
00:06:18.530 --> 00:06:21.970
And we can say yes, the mean value theorem does indeed apply.
00:06:23.050 --> 00:06:30.310
This example demonstrates that we simply need to check the conditions required for the mean value theorem to establish whether we can use it for a given function.
00:06:30.990 --> 00:06:33.440
Now let’s look at an example of how we might use it.
00:06:34.230 --> 00:06:44.890
For the function 𝑓 of 𝑥 equals 𝑥 cubed minus four 𝑥, find all the possible values of 𝑐 that satisfy the mean value theorem over the closed interval negative two to two.
00:06:45.430 --> 00:06:56.570
Remember, the mean value theorem says that if 𝑓 is a function which is continuous over some closed interval 𝑎 to 𝑏 and differentiable at every point of some open interval 𝑎 to 𝑏.
00:06:56.890 --> 00:07:06.260
Then there exists a point 𝑐 in that open interval such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
00:07:06.790 --> 00:07:10.390
Our 𝑓 of 𝑥 is equal to 𝑥 cubed minus four 𝑥.
00:07:10.650 --> 00:07:13.420
And our closed interval is from negative two to two.
00:07:13.790 --> 00:07:22.640
We’re looking to find the value of 𝑐 such that the derivative of our function evaluated at 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
00:07:23.050 --> 00:07:27.440
That’s 𝑓 of two minus 𝑓 of negative two over two minus negative two.
00:07:27.840 --> 00:07:29.080
So we’ll do two things.
00:07:29.170 --> 00:07:30.860
We’ll evaluate this quotient.
00:07:30.990 --> 00:07:34.070
And we’ll also find an expression for the derivative of our function.
00:07:34.500 --> 00:07:37.290
𝑓 of two is two cubed minus four times two.
00:07:37.580 --> 00:07:41.820
And 𝑓 of negative two is negative two cubed minus four times negative two.
00:07:42.210 --> 00:07:44.420
And actually, this gives us a value of zero.
00:07:44.750 --> 00:07:51.330
So in order to find the values for 𝑐 such that 𝑓 prime of 𝑐 is equal to zero, let’s find the derivative of our function.
00:07:51.750 --> 00:07:54.580
The derivative of 𝑥 cubed is three 𝑥 squared.
00:07:54.580 --> 00:07:57.380
And the derivative of negative four 𝑥 is negative four.
00:07:57.670 --> 00:08:00.870
So 𝑓 prime of 𝑥 is three 𝑥 squared minus four.
00:08:01.300 --> 00:08:05.240
And we can say that 𝑓 prime of 𝑐 is equal to three 𝑐 squared minus four.
00:08:05.550 --> 00:08:08.350
So let’s set this equal to zero and solve for 𝑐.
00:08:08.670 --> 00:08:14.190
We begin by adding four to both sides of our equation to obtain that three 𝑐 squared is equal to four.
00:08:14.610 --> 00:08:16.080
Next, we divide by three.
00:08:16.500 --> 00:08:19.260
And we see that 𝑐 squared is equal to four-thirds.
00:08:19.590 --> 00:08:25.520
And finally, we take the square root of both sides, remembering to take by the positive and negative square root of four-thirds.
00:08:25.830 --> 00:08:29.980
To obtain that 𝑐 is equal to plus or minus the square root of four over three.
00:08:30.500 --> 00:08:34.810
And which we can say is equal to two over root three and negative two over root three.
00:08:35.380 --> 00:08:42.580
Notice also that these values of 𝑐 are indeed in the closed interval negative two to two as required by the mean value theorem.
00:08:44.840 --> 00:08:54.930
For the function 𝑓 of 𝑥 equals 𝑥 minus one to the eighth power, find all the possible values of 𝑐 that satisfy the mean value theorem over the closed interval zero to two.
00:08:55.640 --> 00:09:05.930
Remember, the mean value theorem says that if 𝑓 is a function which is continuous over the closed interval 𝑎 to 𝑏 and differentiable at every point of the open interval 𝑎 to 𝑏.
00:09:06.190 --> 00:09:14.430
Then there’s a point 𝑐 in that open interval, such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
00:09:14.950 --> 00:09:17.710
Our 𝑓 of 𝑥 is 𝑥 minus one to the eighth power.
00:09:18.060 --> 00:09:20.700
And our closed interval runs from zero to two.
00:09:21.240 --> 00:09:30.470
We’re looking to find the value of 𝑐 such that the derivative of our function evaluated at 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
00:09:30.980 --> 00:09:34.600
That’s 𝑓 of two minus 𝑓 of zero over two minus zero.
00:09:34.920 --> 00:09:36.020
So we’ll do two things.
00:09:36.020 --> 00:09:37.510
We’ll evaluate this quotient.
00:09:37.510 --> 00:09:42.150
And we’ll also find an expression for the derivative of our function and evaluate it at 𝑐.
00:09:42.570 --> 00:09:48.730
𝑓 of two minus 𝑓 of zero is two minus one to the eighth power minus zero minus one to the eighth power.
00:09:49.020 --> 00:09:55.140
And we obtain that 𝑓 of two minus 𝑓 of zero over two minus zero is zero divided by two, which is just zero.
00:09:55.600 --> 00:09:58.410
Our next job is to find the derivative of our function.
00:09:58.820 --> 00:10:00.910
And we’ll use the general power rule.
00:10:01.080 --> 00:10:09.610
This says that if 𝑔 of 𝑥 is a differentiable function and 𝑛 is 𝑎 constant real number, such that 𝑓 of 𝑥 is equal to 𝑔 of 𝑥 to the power of 𝑛.
00:10:09.850 --> 00:10:18.740
Then the derivative of 𝑓 of 𝑥, 𝑓 prime of 𝑥, is equal to 𝑛 times 𝑔 of 𝑥 to the power of 𝑛 minus one times the derivative of 𝑔 of 𝑥.
00:10:18.740 --> 00:10:19.920
That’s 𝑔 prime of 𝑥.
00:10:20.180 --> 00:10:30.140
The derivative of our function 𝑥 minus one to the eighth power is therefore eight times 𝑥 minus one to the seventh power multiplied by the derivative of 𝑥 minus one, which is one.
00:10:30.540 --> 00:10:34.810
And so we see that 𝑓 prime of 𝑥 is eight times 𝑥 minus one to the seventh power.
00:10:35.240 --> 00:10:39.450
We see that 𝑓 prime of 𝑐 is eight times 𝑐 minus one to the seventh power.
00:10:39.910 --> 00:10:43.860
And remember, we found that 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎 is zero.
00:10:44.100 --> 00:10:47.600
So we’re going to set this equal to zero and then solve for 𝑐.
00:10:48.000 --> 00:10:49.250
We divide by eight.
00:10:49.250 --> 00:10:51.950
And we see that 𝑐 minus one to the seventh power is zero.
00:10:52.270 --> 00:11:02.170
And we can take the seventh root of both sides to see that 𝑐 minus one is equal to zero, which means that 𝑐 equals one is the only value of 𝑐 that satisfies the mean value theorem.
00:11:02.400 --> 00:11:05.800
Notice it falls in the closed interval zero to two, as required.
00:11:07.030 --> 00:11:11.580
In our final example, we’ll look at how this theorem can be applied to a contextual question.
00:11:12.830 --> 00:11:15.130
A rock is dropped from a height of 81 feet.
00:11:15.520 --> 00:11:24.480
Its position 𝑡 seconds after it is dropped until it hits the ground is given by the function 𝑠 of 𝑡 equals negative 16𝑡 squared plus 81.
00:11:25.140 --> 00:11:27.490
Determine how long it will take for the rock to hit the ground.
00:11:28.060 --> 00:11:31.860
Find the average velocity of the rock from the point of release until it hits the ground.
00:11:32.240 --> 00:11:38.920
And find the time 𝑡 according to the mean value theorem when the instantaneous velocity of the rock is equal to the average velocity.
00:11:39.380 --> 00:11:44.590
The rock will reach the ground when its position 𝑠 of 𝑡 is equal to zero.
00:11:44.880 --> 00:11:51.600
We can therefore set this expression negative 16𝑡 squared plus 81 equal to zero and solve for 𝑡.
00:11:51.950 --> 00:11:56.800
We add 16𝑡 squared to both sides and then divide through by 16.
00:11:56.800 --> 00:12:00.480
And we obtain 𝑡 squared to be equal to 81 over 16.
00:12:00.820 --> 00:12:07.040
We then take the square root of both sides, remembering to take by the positive and negative square root of 81 over 16.
00:12:07.370 --> 00:12:11.320
And we see that 𝑡 is equal to plus or minus nine-quarters.
00:12:11.510 --> 00:12:15.840
Now, we can actually disregard negative nine-quarters since we’re working in time.
00:12:16.190 --> 00:12:20.040
And we find that the rock hits the ground after nine-quarters of a second.
00:12:20.560 --> 00:12:24.960
Our next job is to find the average velocity of the rock over this period of time.
00:12:25.290 --> 00:12:29.550
The definition for average velocity is total displacement divided by time taken.
00:12:29.830 --> 00:12:32.490
The displacement of our rock is its change in position.
00:12:32.490 --> 00:12:34.310
That’s negative 81 feet.
00:12:34.630 --> 00:12:37.390
And it takes nine-quarters of a second to travel this far.
00:12:37.600 --> 00:12:41.140
So the velocity is negative 81 divided by nine over four.
00:12:41.510 --> 00:12:46.020
Remember, to divide by a fraction, we can multiply by the reciprocal of that fraction.
00:12:46.220 --> 00:12:49.020
So we have negative 81 times four over nine.
00:12:49.390 --> 00:12:51.230
And then we cancel this factor of nine.
00:12:51.660 --> 00:12:56.000
And so we obtain that the average velocity of our rock is negative 36 feet per second.
00:12:56.680 --> 00:13:00.050
For the final part of this question, we’ll need to quote the mean value theorem.
00:13:00.320 --> 00:13:09.410
Remember, this says that if 𝑓 is a continuous function over some closed interval 𝑎 to 𝑏 and differentiable at every point of that open interval 𝑎 to 𝑏.
00:13:09.650 --> 00:13:17.520
Then there’s a point 𝑐 in this interval, such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.
00:13:17.890 --> 00:13:23.240
We know that the average velocity is negative 36 feet per second.
00:13:23.520 --> 00:13:26.040
That’s the equivalent of this quotient.
00:13:26.470 --> 00:13:32.230
The instantaneous velocity can be found by differentiating our function for position.
00:13:32.590 --> 00:13:36.150
That’s 𝑠 prime of 𝑡 equals negative 32𝑡.
00:13:36.550 --> 00:13:41.940
In this case then, we can say that 𝑠 prime of 𝑐 is equal to negative 32𝑐.
00:13:42.120 --> 00:13:46.450
And we obtain the equation negative 32𝑐 equals negative 36.
00:13:46.740 --> 00:13:50.390
We solve for 𝑐 by dividing both sides by negative 32.
00:13:50.770 --> 00:13:59.450
And we find that the time at which the instantaneous velocity of the rock is equal to the average velocity is equal to nine-eighths of a second.
00:14:00.530 --> 00:14:03.100
In this video, we briefly discussed Rolle’s theorem.
00:14:03.470 --> 00:14:07.570
And we said that if a function 𝑓 satisfies three criteria.
00:14:07.840 --> 00:14:15.630
That is, it’s continuous over the closed interval 𝑎 to 𝑏, differentiable on the open interval 𝑎 to 𝑏, and 𝑓 of 𝑎 equals 𝑓 of 𝑏.
00:14:15.920 --> 00:14:23.290
Then, there exists a number 𝑐 in that open interval such that the derivative of 𝑓 evaluated at 𝑐 is equal to zero.
00:14:23.560 --> 00:14:27.070
We also saw that we can use Rolle’s theorem to prove the mean value theorem.
00:14:27.580 --> 00:14:34.470
And this says that if 𝑓 of 𝑥 is continuous over a closed interval 𝑎 to 𝑏 and differentiable over an open interval 𝑎 to 𝑏.
00:14:34.680 --> 00:14:43.740
Then there exists some number 𝑐 in that open interval such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎.