Video: Finding the Local Maximum and Minimum Values of a Polynomial Function

Find the local maximum and minimum values of 𝑓(π‘₯) = βˆ’5π‘₯ + (π‘₯ βˆ’ 2)⁡ βˆ’ 5.

03:43

Video Transcript

Find the local maximum and minimum values of 𝑓 of π‘₯ equals negative five π‘₯ plus π‘₯ minus two to the fifth power minus five.

Local minima and maxima are examples of critical points of a function. And of course, we know that critical points occur when the first derivative is equal to zero or possibly does not exist. And there are a couple of ways we can establish whether these are maximums or minimums. One of these method is to use the second derivative test. We evaluate the second derivative at each point, and if the second derivative is negative, we know we have a local maximum. And if it’s positive, we have a local minimum. And so, it should be quite clear we’re going to need to begin by differentiating 𝑓 of π‘₯. We can do this term by term.

The derivative of negative five π‘₯ is quite straightforward. It’s negative five. Then, we can use the general power rule to differentiate π‘₯ minus two to the fifth power. Remember, this is just a special case of the chain rule. We multiply the entire term by the exponent and reduce the exponent by one. And then, we multiply that expression by the derivative of the inner function. Of course, the derivative of π‘₯ minus two is simply one. So, we get five times π‘₯ minus two to the fourth power times one, which is simply five times π‘₯ minus two to the fourth power. And the derivative of a constant is zero. So, we’ve done. We’ve found the first derivative. It’s negative five plus five times π‘₯ minus two to the fourth power.

So, let’s set this equal to zero and solve for π‘₯. We’ll begin by adding five to both sides of the equation. We find that five is equal to five times π‘₯ minus two to the fourth power. We’ll then divide through by five, and we find that one is equal to π‘₯ minus two to the fourth power. Next, we address the fourth power. 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. So, π‘₯ minus two must be equal to either positive one or negative one. Let’s deal with the equation that says π‘₯ minus two is equal to positive one.

We solve that by adding two to both sides, and we find that π‘₯ is equal to three. And when π‘₯ minus two is equal to negative one, we still add two to both sides. But this time, we find that π‘₯ is equal to one. So, we have some critical points when π‘₯ is equal to three and π‘₯ is equal to one. But we’re going to need to differentiate our expression again to work out whether these are maximums or minimums. Let’s begin by differentiating negative five.

Now, of course, the derivative of a constant is zero. We use the general power rule again to differentiate five times π‘₯ minus two to the fourth power with respect to π‘₯. We multiply the entire term by four. That gives us four times five which is 20. Then, we reduce the exponent by one, so we get π‘₯ minus two to the third power. And then, we multiply this by the derivative of the inner function, but that’s just one. So, 𝑓 double prime of π‘₯, the second derivative with respect to π‘₯, is 20 times π‘₯ minus two cubed. Our job now is to establish whether the second derivative at π‘₯ equals three and the second derivative at π‘₯ equals one are local minima or local maxima.

When π‘₯ is equal to three, we get 20 times three minus two cubed which is 20, which is positive. When π‘₯ is equal to one, we get 20 times one minus two cubed, which is negative 20. And that’s negative. And so, we know when π‘₯ is equal to three, we have a local minimum. And when π‘₯ is equal to one, we have a local maximum. And of course, we could actually have established this by substituting these values into our function. Let’s do that now. When π‘₯ 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. And when π‘₯ is equal to one, we have negative five times one plus one minus two to the fifth power minus five, which is negative 11.

Now, we saw obviously that π‘₯ equals three and π‘₯ equals one were the only critical points we were looking at. And we’ve seen that the value of the function at three is less than the value of the function at one. And so, we’ve established that the function 𝑓 of π‘₯ equals negative five π‘₯ plus π‘₯ minus two to the fifth power minus five has a local maximum of negative 11 at π‘₯ equals one and a local minimum of negative 19 at π‘₯ equals three.

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