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### Problem 9

02:35
Georgia Southern University

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Problem 8

$$\begin{array}{l}{\text { a. What are the critical points of } f ?} \\ {\text { b. On what open intervals is } f \text { increasing or decreasing? }} \\ {\text { c. At what points, if any, does } f \text { assume local maximum and }} \\ \quad {\text { minimum values? }}\end{array}$$
$$f^{\prime}(x)=\frac{(x-2)(x+4)}{(x+1)(x-3)}, \quad x \neq-1,3$$

$$\begin{array}{l}{\text { (a) Critical points } x=-4,-1,2,3 ](b)[\text { see the graph }]} \\ {\text { (c) local maximmm point } x=-4,2 \text { local minimum point } x=}\end{array}$$

## Discussion

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## Video Transcript

Okay, so I drew It is this time is a rational function, not just a polynomial Teo X minus, two times X plus for and then x plus one x minus three. And so we have that X cannot be equal to negative one or three course. So if you look at our critical points first thing to Dio so we see that X equals two will make of prime zero the numerator zero and then X equals negative for and then the derivative can be undefined. So that's when exes negative one and X is three. But now these are these are excluded from the domain of our function raises their ass and toads. So we should still think about what's happening to the left and the right and those guys. So if we look, let's plot all of these numbers. So he has do you for one, two, three. And we want to look at f prime because that's going to tell us where death is increasing or decreasing. We're privates, positive or negative. So if f if X is less than negative for this is going to be negative, negative maga tive and night. So we're gonna have four negatives all being multiplied by together. Says going given met positive. Good. And then between negative form one. Well, there are going to be negative, but then this term is going to become positive. So we're only gonna have three. Agnes, this could be negative. Negative, negative, negative, positive mix. And that negative And then between negative one and two, this term becomes positive. So it's just negative. Negative, Positive, Positive. So in that positive and then between two and three, this term becomes positive. Positive, positive, positive. Negative Makes a negative. And in greater than three, they're all positive. Okay, good. So f is increasing than t three saying the increasing and baker saying on in Greece. Okay. And so we are going to be increasing from negative infinity tonight it for decreasing from negative for the ninety one, increasing from negative one to two, decreasing from two to three and then increasing from three to infinity. Great. And we want to look at our local extreme of but we've got to be careful because X equals one and X equals X equals negative one x three are not actually points in the domain of the function so we can't actually have extreme in there, Not the function can change from being increasing, decreasing at across his ass and tits, but they're not gonna actually be extreme. So what are our extreme? A We're gonna have extreme at two. Negative for Looks tough. So it too. All right, let's see. At negative four were increasing and then decreasing. So that's a local Max. Okay? And at two, we're going from increasing, decreasing. So we have another local Max Mexico's still and negative one and three are not actually local minima because those X values air not in the domain of our original function.