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(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 inforvation from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one.$h(x)=(x+1)^{5}-5 x-2$

A. increase: $x<-2$ and $x>0$decrease: $-2< x <0$B. Maximum $-f(-2)=7$Minimum $-f(0)=-1$C. concave up: $x>-1$concave down: $x<-1$inflection points: $x=-1$D. SEE GRAPH

Calculus 1 / AB

Chapter 4

APPLICATIONS OF DIFFERENTIATION

Section 3

Derivatives and the Shapes of Graphs

Derivatives

Differentiation

Applications of the Derivative

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for this question, we can ride out the narrative switch at the second duty of self age, like me. So for party, um, we're looking for the humor increasing decreasing intervals. So it's set a crime. X equals zero. That means we have, um the terms in this bracket equals zero, which gives us X plus one to the fourth Power Coast. One Nahmias X plus one equals two plus or minus one. So we have two solutions. X equals two minus two and X equals zero Eso 37 troubles from my A neg. Negative infinity of connective too connective 2 to 0 in the front, zero to infinity on the first interval Which prime? Miss positive. So the function is increasing on the second over the H promise. Negative. So the functions decrease you on the last interview over the age promised positive. So the phone she is increasing again. And from here we know there is a local so increasing first, then the questions. There's a local next man at X equals two minus two with value um F minus also H minus two recourse to seven and we have a local minimum at X equals 20 with valuable on HDL H zero, you host a minus one. Okay, so for processing, we just said, Let's take on the narrative. You crossed zero. So we have only one solution, which is X equals two minus one. So we have to, as I've been told us from connective infinity connective one connected for into infinity over the first integral H double. Prime Minister Negative said can kept on over a second home of the H. The premier's positive so the functions can keep up. No, we are ready to sketch the graph or fetch. Um, so you really labels some critical point? An inflection point. So from here, we know there is the inflection point at X equals the minus one. Okay, so the function is increasing and the concrete down here on decreasing Conch it down. No, there is, uh, there's an inflection point here. So he changes it contributed to be Come give up. Don't make this. And the way have we have to, um no co it's treacherous. So local minimum at X equals zero in a local maximum and exit post minus two

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