WEBVTT 00:00:01.100 --> 00:00:09.310 Find the local maximum and minimum values of 𝑓 of 𝑥 equals negative five 𝑥 plus 𝑥 minus two to the fifth power minus five. 00:00:09.710 --> 00:00:14.230 Local minima and maxima are examples of critical points of a function. 00:00:14.520 --> 00:00:20.010 And of course, we know that critical points occur when the first derivative is equal to zero or possibly does not exist. 00:00:20.180 --> 00:00:24.720 And there are a couple of ways we can establish whether these are maximums or minimums. 00:00:24.820 --> 00:00:27.800 One of these method is to use the second derivative test. 00:00:27.870 --> 00:00:33.970 We evaluate the second derivative at each point, and if the second derivative is negative, we know we have a local maximum. 00:00:34.260 --> 00:00:35.970 And if it’s positive, we have a local minimum. 00:00:36.080 --> 00:00:40.450 And so, it should be quite clear we’re going to need to begin by differentiating 𝑓 of 𝑥. 00:00:40.520 --> 00:00:42.160 We can do this term by term. 00:00:42.350 --> 00:00:45.100 The derivative of negative five 𝑥 is quite straightforward. 00:00:45.100 --> 00:00:45.950 It’s negative five. 00:00:46.110 --> 00:00:49.900 Then, we can use the general power rule to differentiate 𝑥 minus two to the fifth power. 00:00:50.110 --> 00:00:52.520 Remember, this is just a special case of the chain rule. 00:00:52.620 --> 00:00:56.870 We multiply the entire term by the exponent and reduce the exponent by one. 00:00:57.260 --> 00:01:01.290 And then, we multiply that expression by the derivative of the inner function. 00:01:01.320 --> 00:01:03.870 Of course, the derivative of 𝑥 minus two is simply one. 00:01:03.870 --> 00:01:09.990 So, we get five times 𝑥 minus two to the fourth power times one, which is simply five times 𝑥 minus two to the fourth power. 00:01:10.210 --> 00:01:12.230 And the derivative of a constant is zero. 00:01:12.230 --> 00:01:12.800 So, we’ve done. 00:01:12.800 --> 00:01:14.070 We’ve found the first derivative. 00:01:14.210 --> 00:01:17.850 It’s negative five plus five times 𝑥 minus two to the fourth power. 00:01:18.150 --> 00:01:21.380 So, let’s set this equal to zero and solve for 𝑥. 00:01:21.470 --> 00:01:24.360 We’ll begin by adding five to both sides of the equation. 00:01:24.360 --> 00:01:28.040 We find that five is equal to five times 𝑥 minus two to the fourth power. 00:01:28.220 --> 00:01:32.500 We’ll then divide through by five, and we find that one is equal to 𝑥 minus two to the fourth power. 00:01:32.760 --> 00:01:34.360 Next, we address the fourth power. 00:01:34.430 --> 00:01:42.270 Well, we’re going to take the fourth root of both sides of our equation, remembering that since the exponent is even, we need both the positive and negative fourth root of one. 00:01:42.450 --> 00:01:45.950 So, 𝑥 minus two must be equal to either positive one or negative one. 00:01:46.150 --> 00:01:49.360 Let’s deal with the equation that says 𝑥 minus two is equal to positive one. 00:01:49.490 --> 00:01:53.240 We solve that by adding two to both sides, and we find that 𝑥 is equal to three. 00:01:53.450 --> 00:01:56.860 And when 𝑥 minus two is equal to negative one, we still add two to both sides. 00:01:56.860 --> 00:01:58.780 But this time, we find that 𝑥 is equal to one. 00:01:58.970 --> 00:02:03.100 So, we have some critical points when 𝑥 is equal to three and 𝑥 is equal to one. 00:02:03.310 --> 00:02:08.230 But we’re going to need to differentiate our expression again to work out whether these are maximums or minimums. 00:02:08.360 --> 00:02:10.550 Let’s begin by differentiating negative five. 00:02:10.620 --> 00:02:13.010 Now, of course, the derivative of a constant is zero. 00:02:13.230 --> 00:02:18.930 We use the general power rule again to differentiate five times 𝑥 minus two to the fourth power with respect to 𝑥. 00:02:18.980 --> 00:02:21.120 We multiply the entire term by four. 00:02:21.150 --> 00:02:22.980 That gives us four times five which is 20. 00:02:23.350 --> 00:02:26.660 Then, we reduce the exponent by one, so we get 𝑥 minus two to the third power. 00:02:26.870 --> 00:02:30.170 And then, we multiply this by the derivative of the inner function, but that’s just one. 00:02:30.580 --> 00:02:36.280 So, 𝑓 double prime of 𝑥, the second derivative with respect to 𝑥, is 20 times 𝑥 minus two cubed. 00:02:36.400 --> 00:02:44.920 Our job now is to establish whether the second derivative at 𝑥 equals three and the second derivative at 𝑥 equals one are local minima or local maxima. 00:02:45.150 --> 00:02:50.170 When 𝑥 is equal to three, we get 20 times three minus two cubed which is 20, which is positive. 00:02:50.240 --> 00:02:54.550 When 𝑥 is equal to one, we get 20 times one minus two cubed, which is negative 20. 00:02:54.580 --> 00:02:55.480 And that’s negative. 00:02:55.580 --> 00:02:58.180 And so, we know when 𝑥 is equal to three, we have a local minimum. 00:02:58.420 --> 00:03:00.710 And when 𝑥 is equal to one, we have a local maximum. 00:03:00.850 --> 00:03:05.330 And of course, we could actually have established this by substituting these values into our function. 00:03:05.540 --> 00:03:06.410 Let’s do that now. 00:03:06.590 --> 00:03:13.840 When 𝑥 is equal to three, our function is negative five times three plus three minus two to the fifth power minus five, which is negative 19. 00:03:13.980 --> 00:03:20.320 And when 𝑥 is equal to one, we have negative five times one plus one minus two to the fifth power minus five, which is negative 11. 00:03:20.360 --> 00:03:24.770 Now, we saw obviously that 𝑥 equals three and 𝑥 equals one were the only critical points we were looking at. 00:03:24.910 --> 00:03:29.230 And we’ve seen that the value of the function at three is less than the value of the function at one. 00:03:29.500 --> 00:03:42.630 And so, we’ve established that the function 𝑓 of 𝑥 equals negative five 𝑥 plus 𝑥 minus two to the fifth power minus five has a local maximum of negative 11 at 𝑥 equals one and a local minimum of negative 19 at 𝑥 equals three.