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University of California, Riverside

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Problem 7

Answer the following questions about the functions whose derivatives are given:

\begin{equation}\begin{array}{l}{\text { a. What are the critical points of } f ?} \\ {\text { b. On what open intervals is } f \text { increasing or decreasing? }} \\ {\text { c. At what points, if any, does } f \text { assume local maximum and }} \\ \quad {\text { minimum values? }}\end{array}\end{equation}

\begin{equation}f^{\prime}(x)=\frac{x^{2}(x-1)}{x+2}, \quad x \neq-2\end{equation}

Answer

$$

-2,0,1

$$

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## Discussion

## Video Transcript

Okay, so for problem seven were given a fraction. We chased that. The room too. Attraction is X squared. Times X minus one, divided by X plus two and X is none. X is not equal Thio negative too. Okay, So our first party in this problem is to find the critical points. Now, to this fraction, we just we only need to make our numerator to be zero. So that is acts to be zero or X is one. So these two are our Rico points now, Part B. We need to find a by the increasing the rose and increasing levels. So that is still we need to let destruction to be hostage as salt for the inequality. Now, notice that, um in this case, X X squared is always a negative. So to be to make this whole this whole direction, to be positive, our ex cannot be zero. Otherwise, our our entire direction will be zero. And at the same time, we need to make a fraction X minus one and express to to be to be positive. So that is same to say, Ex tremendous one is positive, and X plus two is positive or we can make them both connective. We stood s steel works because connective number divided by a naked number will be positive. So or we can make explain is one to be negative and x plus two to be negative. Okay. Now, by solving these sets off in qualities, what we have is ex bigger than one or sale or here or acts is smaller than connected to. So that means from negative infinity to connect it to union one to infinity dysfunction will increase and from next 2 to 1, dysfunction will decrease. Okay, Now, for part C, our local maximum will be away on. So yeah, it's doing this problem. We don't have any look Mexico. But we have a local minima because from connected 2 to 1 from the function will decrease now from one to infinity function will increase. So at axes, axes equal to one, we can reach the know Camino and no local maximum

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