Question Video: Finding the Absolute Maximum and Minimum Values of a Polynomial Function on a Closed Interval | Nagwa Question Video: Finding the Absolute Maximum and Minimum Values of a Polynomial Function on a Closed Interval | Nagwa

Question Video: Finding the Absolute Maximum and Minimum Values of a Polynomial Function on a Closed Interval Mathematics

Determine the absolute maximum and minimum values of the function 𝑓(𝑥) = 2𝑥⁴ − 8𝑥² − 13 in the interval [−1, 2].

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Video Transcript

Determine the absolute maximum and minimum values of the function 𝑓 of 𝑥 equals two 𝑥 to the fourth power minus eight 𝑥 squared minus 13 in the closed interval negative one to two.

Remember, to find absolute extrema for some continuous function 𝑓 of 𝑥, we follow three steps. We start by finding all critical points in the closed interval. We check the values of 𝑓 of 𝑥 at these critical points. And then, we check the end points for any absolute extrema, values that might be smaller than any relative minima or larger than any relative maxima. Critical points occur when the derivative of the function is equal to zero, or does not exist. So, let’s begin by finding the derivative of our function.

The derivative of 𝑓 of 𝑥 is written as 𝑓 prime of 𝑥, and it’s four times two 𝑥 to the third power minus two times eight 𝑥. And, of course, the derivative of negative 13 is zero. So, we see that the derivative of our function is eight 𝑥 cubed minus 16𝑥. We’ll set this equal to zero and solve for 𝑥. Here, we factor the expression on the right-hand side to get eight 𝑥 times 𝑥 squared minus two.

And we see that for eight 𝑥 times 𝑥 squared minus two to be equal to zero, either eight 𝑥 must be equal to zero, which means 𝑥 is equal to zero. Or we can say that 𝑥 squared minus two must be equal to zero. And solving this, we see that 𝑥 is equal to both the positive and negative square root of two. So, we have critical points on our function at 𝑥 equals zero, 𝑥 equals negative root two, and 𝑥 equals root two.

Our second step is to evaluate the function at these critical points. That’s 𝑓 of zero, 𝑓 of root two, and 𝑓 of negative two. 𝑓 of zero is two times zero to the fourth power minus eight times zero squared minus 13, which is negative 13. 𝑓 of root two is two root two to the fourth power minus eight root two squared minus 13, which is negative 21. And, actually, 𝑓 of negative root two is also negative 21. Now this by itself doesn’t help us much. It certainly looks like negative 21 and negative 13 might be local extrema, but we need to know whether they are absolute extrema.

So, we’re going to evaluate our function at the ends of our interval. Remember, this is because we know that if we have a continuous function over some closed interval 𝑎 to 𝑏, then we’re guaranteed at some point in this interval to have both an absolute maximum and an absolute minimum. And these extreme values are obtained either at the place or places where local extrema occur, or at the end points of the interval.

So, we’re going to evaluate 𝑓 of negative one, and 𝑓 of two. 𝑓 of negative one is two times negative one to the fourth power minus eight times negative one squared minus 13, which is negative 19. And 𝑓 of two is two times two to the fourth power minus eight times two squared minus 13, which is negative 13. We can quite clearly see then that the absolute maximum value of our function in the closed interval negative one to two is negative 13. And the absolute minimum value is negative 21. This example involves some fairly simple differentiation, so we’ll now look at an example which involves a little more work.

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