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03:06 University of California, Riverside
Problem 5

# 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)=(x-1)(x+2)(x-3)\end{equation}

## Discussion

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## Video Transcript

Kate. So four problem five were given that the roof too. That iss X minus one Experts too. And times X minus three. All right, so what we need to do is first find Theo Critical points. That is, when next, a sequel to one and access equal to negative too. And access equal to three. And any of these will make our directive to be equal to zero. So these are the art of critical points. Now, for our king, you need to find the increase interval and open on dhe decrease interval. So what we need to do first less lead is derivative to be positive, Kate. So four problem five were given that the roof too. That is X minus one. Experts too. And times X minus three. All right, So what we need to do is first find Thea. Now, the only thing we need to do is to solve this sub Vizzini equality. Now they are 22 conditions. We need to consider here the first condition ISS. All these three terms are all positive. So that is to say, our expert minus one is positive and article points. That is when next a sequel to one and access equal to negative too, and access equal to three. And any of these will make our directive to be equal to zero. So these are the art of critical points. Now are you need to find the increase interval and open on dhe decrease interval. So what we need to do first less. Let is derivative to be positive or X plus two is positive and X minus three. It's positive and combine these two these three inequalities then we have. So what we have will be X B bigger than three, and the second condition we need to consider is to have two negative things that is now. The only thing we need to do is to solve this sub Vizzini equality. Now they are 22 conditions that we need to consider here the first condition ISS. All these three terms are all positive. So that is to say, our expert minus one is positive and our either x one x minus. Well experts, too, are negative or X plus two plus express to an X minus. Three are negative or X minus one and x minus. Three are negative. So that is we have X minus one. Uh, I'll start a new page, so our condition, too, will be X plus two is positive and X minus three. It's faster and combine these two these three inequalities then we have. So what we have will be X B bigger than three, and the second condition we need to consider is too have to negative things. That is two negative terms. Okay, so we to neck to turn to connect the terms. Let's first consider X minus one is next to and X plus two. It's not, too, and X minus three is positive. We cannot let the third time either x one x minus. While experts, too, are negative or experts too, plus express to an X minus, three are negative or X minus one and X minus. Three are negative. So that is we have X minus one. Uh, I'll start a new page so our condition, too, will be to connective. Otherwise, solver are a whole equation will be connected. So this is the first condition. Or we could have the 2nd 2nd of sorry for this is the first inequality. Or we could have the second inequality which is make explains one neck, too. And X plus two house, too and X minus three. Two negative terms. Okay, so we too negative term to connect the terms nest first, consider X minus one is next to and X plus two. It's not, too, and X minus three is past it. We cannot let the third term to pit or the last the last in qualities we need to think about. ISS X minus one wants to and X plus two negative and X minus tree naked. So by solving these three set, it's three sets of inequalities. What we have in the end is connective. Otherwise, solver are a whole equation will be connected. So this is the first condition. Or we could have the 2nd 2nd of the purchases. The first the inequality. Or we can have the second inequality, which is make explains one next to and X plus two house too and X minus three night, um, empty for the first set of inequalities and negative too smaller than acts smaller than one. It's our second date. Quality is a solution of our over a seven second set of inequalities and the third in qualities or so gives us. Um Empty said so Combine one and two or the last. The last in qualities we need to think about it is X minus. One wants to and X plus two negative and X minus tree Noted So by solving these three set is three sets of inequalities. What we have in the end is due. We conclude that on interval from negative too from one union, three from infinity dysfunction will increase and from negative that could be penetrated to connect it to you again. 123 dysfunction will decrease. Okay, now, R. C, we want to find a local, make Mexico the local minimum. So Daddy's just till, um, empty for the first set of inequalities and negative too smaller than acts smaller than one. It's our second date. Quality is a solution of our of our set a second set of inequalities and the third in qualities or so gives us, um empty said so Combine one and to some of these, uh, these two in a boat here. So first we reach our from from negative 2 to 1. We reach our local Mexico because the function is increasing. And after one it will decrease from 1 to 3. So that one will be loco will be a local Mexico and our local minimum The officer from Active Benedetti connected to the function This always increase, always decreasing you. We conclude that on interval from negative too from one union three wrong infinity dysfunction will increase and from negative that could be penetrated to connect it to you again. 123 dysfunction will decrease. Okay now R c. We want to find a local make Mexico the local minimum. So Daddy's just the opposite and from neck to 2 to 1 to function were increased, So elected to will be local minimum and also from 1 to 3 the function decrease and from three to infinity dysfunction will increase. So f three we also be a local minimal so
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