a. Identify the function's local extreme values in the given domain, and say where they occur.

b. Which of the extreme values, if any, are absolute?

c. Support your findings with a graphing calculator or computer grapher.

$$

f(x)=2 x-x^{2}, \quad-\infty<x \leq 2

$$

## Discussion

## Video Transcript

All right. So here are function is X good X cubed over three X squared plus one. And if we take the derivative, this is going to be a quotient. So we have three x squared plus one times through X squared minus X cubed times six x and all over three X Square plus one squared. So the denominator here is always strictly greater than zero. So the drug is never going to be on to find. But if we want to know where f crime is zero, we just want to set. Okay, let's multiply this out. We have nine next to the fourth plus three x squared minus six x to the fourth zero. Or, in other words, three x to the fourth equals three x in the fourth plus three x squared zero. So in other words, three x squared times x squared plus one zero. But this is only zero and exits here because X squared plus one is always positive. Say it one critical one zero. So if you look yes, the important thing notices that maybe it's not clear but the okay, So the denominator of the derivative is his positive and the numerator of the drift was positive. The numerator is right here. It's three extra, the fourth most three x squared. That's all. Even powers of X. That's all positive. So the derivative is always greater than or equal to zero. So we're going to be positive here and positive here. So then F is going to be increasing from negative infinity to zero and increasing from zero to infinity. But then this function is just straight increasing all the way through. And there's no local maximum amendment never changes from being increasing to decreasing, so they're no extremely local or absolute.

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