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Numerade Educator



Problem 41 Medium Difficulty

(a) Find the intervals of increase or decrease.
(b) Find the local maximum and minimum values.
(c) Find the intervals of concavity and the inflection points.
(d) Use the information from parts $ (a) - (c) $ to sketch the graph.
Check your work with a graphing device if you have one.

$ h(x) = (x + 1)^5 - 5x - 2 $


(a) $h(x)=(x+1)^{5}-5 x-2 \Rightarrow h^{\prime}(x)=5(x+1)^{4}-5 . h^{\prime}(x)=0 \Leftrightarrow 5(x+1)^{4}=5 \Leftrightarrow(x+1)^{4}=1 \Rightarrow$
$(x+1)^{2}=1 \Rightarrow x+1=1$ or $x+1=-1 \Rightarrow x=0$ or $x=-2 . h^{\prime}(x)>0 \Leftrightarrow x<-2$ or $x>0$ and
$h^{\prime}(x)<0 \Leftrightarrow-2<x<0 .$ So $h$ is increasing on $(-\infty,-2)$ and $(0, \infty)$ and $h$ is decreasing on (-2,0)
(b) $h(-2)=7$ is a local maximum value and $h(0)=-1$ is a local minimum value.
(c) $h^{\prime \prime}(x)=20(x+1)^{3}=0 \Leftrightarrow x=-1 . \quad h^{\prime \prime}(x)>0 \Leftrightarrow x>-1$ and
$h^{\prime \prime}(x)<0 \Leftrightarrow x<-1,$ so $h$ is $\mathrm{CU}$ on $(-1, \infty)$ and $h$ is $\mathrm{CD}$ on $(-\infty,-1)$
There is a point of inflection at $(-1, h(-1))=(-1,3)$


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Video Transcript

this problem. We're going to be analyzing the graph by looking at the function. This is where derivatives and second derivatives could be really helpful because they allow us to understand how graphs are shaped. So the graph that were, um, or gonna be graphing is X plus one of the fifth, and that will be plus or minus five X minus two. So this is the graph that we are dealing with. It is the more interesting part of the graph. Then we want to determine the intervals of increase in decrease. So it's increasing from negative infinity too negative, too decreasing from negative 2 to 0 and then increasing from zero to infinity, the local minimums and maximums. We can look at the graph or we can also look at the derivative of the graph. We see that there's a local maximum at negative two seven and a local minimum at zero and zero negative one. So with all this in mind, we also want to look at con cavity. So here's the con cavity graph. We see that there is an inflection point at negative one. Um, and this is because the graph is con cave down from negative infinity to negative one. And then Khan gave up from negative one to infinity. So with all this information that we see from the derivative and the second relative were able to grab our function which will look something like this