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Problem 12

Answer the following questions about the function…

Problem 11

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)=x^{-1 / 3}(x+2)


\begin{array}{l}{\text { (a) Critical points } x=-2,0 |(b) \text { see the graph }} \\ {\text { (c) local maximum point } x=-2 \text { local minimum point } \quad x=0}\end{array}



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Video Transcript

It's so for 11. Now we have the derivative, which is X to the power off. Negative 1/3 time's X plus two. Now for part A. We just need to find a critical points. Now, in this case, we let this dream team to be equal to zero. Now, acts can be zero, or acts can be connected to get now for part B. That's led this directive to be positive now to solve this inequality. Um, so the first thing we can do is too assumed that now we have two conditions to go to consider here that is exposed to and X are all positive, or we can have experts to. And acts are all negative because to the product of two, negative numbers will be positive. So, uh, second manager, will you bastard, positive. They're real too. Okay, now, okay. From the first set of inequalities, we can. Now we can we can find from this diversity inequality experts to his bigger than zero. So actually, speaker, they're connected to and intersected the positive act so we can concludes that xB actually speaker than zero. Okay, Now, for the 2nd 2nd set up inequalities, we first have express to is smaller than zero, so act is smaller than a negative to now intersect the excess smaller, then smaller than zero. So our second, our solution for the second quality will be excess is smaller than collective too. So that means from next negative infinity to elective to union zero to infinity, this function will be increasing. Otherwise it will be depressing. That is, from next 2 to 0. For Parsi, find a local maximum. We just observed this interval because from negative negative infinity to elected to this function is increased and from negative to zero, the pumpkin will be decreasing. So at this point, active too, our function will reach a local Mexico No CO maximum. Now the local minimum similarity will be x zero because when we reached the zero from Rome next 2 to 0, function will be decreasing and after that function will be increasing. So this will have zero will be a local minimal