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Georgia Southern University

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Problem 10

Answer the following questions about the functions whose derivatives

are given.

a. What are the critical points of $f ?$

b. On what open intervals is $f$ increasing or decreasing?

c. At what points, if any, does $f$ assume local maximum and minimum values?

$$

f^{\prime}(x)=3-\frac{6}{\sqrt{x}}, \quad x \neq 0

$$

Answer

$$

\begin{array}{l}{\text { (a) Critical points } x=0,4} \\ {c )[\text { No local maximum point local minimum point } x=4}\end{array}

$$

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## Discussion

## Video Transcript

Okay, so we know that the derivative of a function f is three minus six over square root of X. And we're told that X not you two zero. All right, So, yeah, when is F prime equal to zero? And when is it undefined? Well, of course it's going to be undefined zero, but F is also under find it. Zero. So we will include that as a critical point, um, to see where the function is increasing or decreasing. So you can actually rewrite this if we put kind of a square root of X. Yeah. So you're all just isn't scratcher clicks a crime of eggs. I can write his three read X minus six. Overruled X And so I'm looking for a value backs that makes this equal to zero. Well, if I put in for I'm going to get square to force to three times two, six, six nine, six zero. So we have a critical point at for. And then, of course, we have this point where the derivative is undefined and also the functions undefined. Okay, so let's think about where the function is increasing or decreasing. Or, in other words, where death is going to be f crime is going to be positive or negative. Well, look at crime. Yeah, well, look a zero for Okay, Okay. So to the left of zero. Well, that's gonna be a problem, right? Because if crime is not defined for ex lessons here because if I try to plug in a number less than zero, I'm going to get, uh, the square root of a native number. So it's really undefined. Not here between zero and four. It's a If I plug in one, I get three minus six over one. That's gonna be a negative number. So f promise going to being negative between zero. And for what? If I plug in something greater than for this numerator is going to be positive over a positive number. So I'm going to get something positive. So then f is going to be decreasing between zero and four and then increasing between for infinity, Okay? And then if I look at my extreme well, zero, it's not going to be anything right, because the function just come on Xueling and redefined for ex greater than zero. So it seems like we just have a local men at X equals war because that's where the function changes from being decreasing to increasing

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