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Show that $5$ is a critical number of the function$$ g(x) = 2 + (x - 5)^3 $$ but $g$ does not have a local extreme value at $5$.

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00:53

Wen Z.

00:17

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 1

Maximum and Minimum Values

Derivatives

Differentiation

Volume

Missouri State University

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

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All right, let's go through this problem we have. Why is equal to two plus X minus five Cube in the question is asking Show that X equals to five gives you the derivative equals to zero. But when I plug in five, why will end up being two and five comments to is not going to be a maximum or minimum. So one of those things that I mentioned that those air called a saddle point. Okay, so let's figure out what why Prime is equal to first two disappears. So this will be three times X minus five quantity squared. The derivative of insight is one, so we don't need to care about that now. If this is equal to zero, it's very easy to calculate that X is equal to five. So that part, I'm sure you can follow so well, like Prime is equal to zero at X equals 25 So why is it that this is not going to be a maximum or minimum? Uh, the precise definition off how ah local maximum and minimum appears Is that the slow pass to be zero off course? Why prime is going to be zero. But the sign has to change at the point, so from negative to positive or positive to negative. So if you look at the graph off three times X minus five, squared at X equals to five. The graph looks roughly like this, and as you can see that the sign doesn't change here. It's positive values here in positive values there, even though why prime is equal to zero. So what happens in the graph off the original one? Let's show you, you can see that it increases over here, but it's still increases there, even though the slope it's momentarily equal to zero at X equals 25 So that's one of the things that could happen just because the derivative is equal to zero. That doesn't necessarily mean that you will have a local maximum or a minimum, and that's how you showed this problem.

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