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Problem 50

a. Identify the function's local extreme values i…

Problem 49

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)=\sqrt{25-x^{2}}, \quad-5 \leq x \leq 5


tee the graph



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Video Transcript

find the local extremes for this function. We can use the chain rule and we'll take the derivative of the outside. Leave the inside alone on this reduces the power down the negative one half, and then we want to multiply by the derivative of the inside, which is negative two X, then setting this equal to zero to find the critical points with an could rearrange this rewrite best getting rid of the twos first, since those cancel will have negative X up top over one half should be a negative one half here, so that means it's over. Um, the fraction and we want to find out when this is equal to zero or when it's undefined. We could see it's equal to zero when x zero, but then it's also going to be undefined when X is either positive or negative. Five sounds we would divide by zero. So we've got three critical points here. We could set up our honorable and then in this problem are Interval does go from negative 5 to 5 and then we have our other critical point so we could test values like these ones in blue. Negative one positive one plug them into this function and see what we get. If we point, then negative one, we get a negative times negative, which would be a positive positive. One would get a negative. So this lets us know that we have a local Max at zero. And then when we plug that back into the top function, we'd end up getting five. And then we also have to local men's on either end point here, and these would be a negative five zero and positive 50 So that's for part A and then part B. We want to graph this so this would look something like this, and all of these would end up being absolute. Max is, um, men's. So he's too, would be these absolute mints. And then we have 05 which would be our absolute max. So it just happens that all of the local maxes amends end up being the absolute Max is, and then

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