University of California, Riverside

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72

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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