a-find-the-intervals-of-increase-or-decrease-b-find-the-local-maximum-and-minimum-values-c-find-t-17

Question

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

Jack Chen · Accepted Answer

for this question, we can ride out the narrative switch at the second duty of self age, like me. So for party, um, we&#x27;re looking for the humor increasing decreasing intervals. So it&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s an inflection point here. So he changes it contributed to be Come give up. Don&#x27;t make this. And the way have we have to, um no co it&#x27;s treacherous. So local minimum at X equals zero in a local maximum and exit post minus two

Carson Merrill · Answer

this problem. We&#x27;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&#x27;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&#x27;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&#x27;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

Jack Chen · Answer

for this question were first right out to the dearer to yourself if for part one we said, I&#x27;ve prompted the zero. So you have X equals zero minus during the one and we can see that the four hour intervals next if infinity to a negative one negative 10 0 to 1 in the one to infinity on the first interval, we can see that if exists, that&#x27;s though that&#x27;s no one. So if crime is negative, the phone is decreasing. On the second her phone negative 1 to 0 If promise qualitative. The function is increasing on the sick on the third interval. If if X is below 0 to 1 if prime is positive, the finishing secrecy If access is great on the what if crime is connective The funny on the Funchess decreasing. So we have a local Oh minima at X equals two minus one with with value F minus one in close to minus two and then we have a local next Men at X equals to one with value if one because the to So this is part true. Have three. We set a second narrative to be zero, so we have X equals zero. Um, I and the process minus one over wrote off to. So we consider the following for intervals minus infinity to minus one over to talk to minus one over it up to two, zero 0 to 1. Over what? Of two at the one over. Root of two to infinity. So we can see the sign of this. So no first interval, we can see that f double promise. Positive. So the functions conflict on the second ago. We can see that if double prime just directive so the functions can keep down on the third interval. If the book crime Yes, positive. So the functions can keep up. And on the last interval of stop a promise next trip, the functions can&#x27;t get them in the Based on this information, we are ready to schedule a graph off. If so, first we laboard&#x27;s on the local minima local maximum. Um, it&#x27;s X equals two minus on the X seaQuest one. So we have minus one in the one said. Of course, bone being bad, it will be minus two and to Okay, so for looking for the inflection point, we have to re inflection point at X equals two finals one over. Ripped off too. And the X equals zero. And that cakes equals to 1/2. Okay, so we are ready to graph. Yeah, relate. So really boards the inflection point by the green. Common. Okay. It&#x27;s a very the graph. So the function is decreasing. First in the conclave up and the increasing Now you become concave down. It did not become Come caper up in the conclave. Dumb again. No decrease in. So this is the grab off function. If so, we have three inflection point. And that one no come makes him and the one no convenient.

Clarissa Noh · Answer

he has close the would name right here. So we&#x27;re gonna first find the derivative of H. We got 15 x square minus 15 X to the fourth. Then we take negative 15 up square. Get X minus one X plus one. So it&#x27;s increasing. Negative one calm. A zero zero comma one and decreasing. Negative. Infinity. Negative one, one comma. Infinity Report. Be so we&#x27;re gonna see that that X is equal to negative one. It goes from decreasing to increasing, so it&#x27;s a minimum and X equals one. It goes from increasing to decreasing, so it&#x27;s a maximum. Now we&#x27;re gonna find the second derivative, which is equal to 30 X minus 60 x cubed. We&#x27;re gonna find where it&#x27;s equal to zero and we get X is equal to zero. X is equal to plus or minus one over square root of two. So we&#x27;re gonna first look up the interval. Negative. Infinity. Pullman Negative One over square root of two. We choose negative one. And what bigger than zero Since h of negative one is bigger than zero, it&#x27;s actually 30. We go negative one over square root of two comma zero. We look at the second derivative of negative 0.1. This is less than zero. You look from zero comma, one over square root of two, and we choose 0.1, which is bigger than zero. And finally we look at one over square root of two to infinity, and we look at one which is less than zero. We clung it in to the regular equation when yet negative 1.237 for each of 00 for H of one over square root of two. That is around 1.237 So we get inflection points to be negative. One over square root of two common negative 1.237 zero comma zero and one of her skirt of two comma 1.237 Next, we&#x27;re gonna draw a quick for us, and it looks something like this

Jack Chen · Answer

okay for this program will first take the purity for Jeep Sergey Prime because to 10/3 extreme minus 1/3 1 minus six So far they at one g prime. Because is there a way have XY costa? But we also need to consider the point. The first the agility of task noti exists so X equals zero. That means we have to raise up intervals to considered from minus infinity to zero from 0 to 1 in the front, one Tween over the first in several key promise less than zero. The function is decreasing over the second car. Virgie prime is politics. The formation is in crazy over the last introvert, You promise? Negative. Again. The function is decreasing. So we have a local minutemen at X equals zero rates very G zero because zero local megs at X equals to one with marriage You want to put the three not for a second of curative the can cavity and the inflection point We take the second narrative which gives us minus 10 overnight Extra de miners for 3rd 1 plus two x No way said she double trying to be zero. We have XY coastal minus 1/2. And then we also need to consider the second narrative test known exists at X equals zero. So from minus infinity is minus 1/2 she&#x27;d have a prime. It&#x27;s politics. So farting is Konkey far from minus one. Have to zero. You have a crime, miss? Um, Conectiv the phone. She&#x27;s conclave down and the from mine was from zero to infinity. You topple premieres connective again. So the funny thing is coming down. Um, from here we have we see that we have to inflection point at X equals two minus one happened X equals zero. Okay, so now we are ready to graph the function on the coordinate. So, firstly, Laboard&#x27;s fashion will pass a long commitment at zero. And the local maximum at X equals to one with very three, and he has to inflection points reaches minus 1/2 and zero. Okay, So when X is greater than zero, we can see that g you see increasing first in a conclave down and the decreasing in the company of thumb. She looks like this. And when exist less than zero, the function is uniformly decreasing. But there&#x27;s an inflection point. So he looks like this. So can cave up. And the conch? A thumb. Um, so there&#x27;s an inflection point here and the here s so we need to change your concave it here so you can&#x27;t give up you on the conclave summer, and it&#x27;s also can get down here.

Jack Chen · Answer

for this question. We have four different parts of in the beginning. We just ride it out of the first so that they were there and then a stakeholder narrative. And so we write out this from solidarity. If I can just do the anemia Factoring, yes, this. So if we write out in this way, it&#x27;s easier to figure out the roots of this if prime. So if the prime that&#x27;s taken out of your day vehicles do minus 12 exits were class four. Okay, so part eight, we&#x27;re going to figure out the increasing and decreasing interval. So you just said if prime of X equals zero so we have three different Rotech Sequels to minus 10 in the one on these three points splits the whole interval, the whole dorm mate with you till four different parts. So the first interface from my a negative infinity tone active one. So over this imperil prime affects, it&#x27;s uniformly positive. We can choose at every point inside this interval in the Prague in tow s prime, and they we can very five Mrs always positive. So it&#x27;s increasing the function. Is the increasing over this in trouble? Next intervals from minus 1 to 0. So I only see terrible. The force of the narrative is negative since Deke Racing, and we do the same thing for the next two from zero to wander from one to positive infinity so we can check off this result by parking some specific value united in trouble. So we have to increasing in Tebow and the toothy crazy interval. Not for part B, part B. We want to find out a local maximum local minimum which is related to so for Patty we know. So this locates the first point. X equals to one X equals minus one. So Onal on the left, it&#x27;s increasing on all the rightist secrets. Meh means we have a local maximum het X equals to one SRE minus one because the increasing for us in decreasing So the value we will be three and we also has also have another local makes that exceed coast. One in the value is also three for the same reason. It&#x27;s increasing for us than the equations. There&#x27;s a peak in a full of local immune. We have one. No commitment attacks equals zero. The value will be too because it&#x27;s decreasing for US 10 million crazy and the for the Integrity Passy connectivity product. We need to use the second intuitive and you just said the second narrative because it&#x27;s euros. So we have X equals two pass minus. Who told 1/3 so have to tweet different points and this pleats little main 23 prods So first product is from minus infinity to minus root off 1/3. So over this interval of the second of curative yes, negative. So it&#x27;s conclave I don&#x27;t and the for a second over from minus root off 1/3 to route all winter A steak on the curative is positive So it&#x27;s conch A farm on the phone for the last one Wrong route off 1/3 toe Positive infinity The second of purity of face Conectiv six come case down again and the foot from here We can see we have to your collection points. Oh, it&#x27;s so why is X minus root off 1/3 And, uh, quality of brutal print there because the continuities um how different at this point? So for off from this information, we can actually sketch the function which is party, so if you want to sketch your function, We need to lay both some specific point, like minus 10 and poor people one and Reza label this minus brutal. Want the other in the root of 1/3 here. So we can see for one and X equals toe one minus one. The value here. It should be three by our argument before. And, uh, the Y intercept is too. So the graph looks like this. So don&#x27;t forget, we have to change it. A cantare ET That&#x27;s miss two points. And we can see we have some critical points year year, and hear off them. Have horizontal tension. Lies. Um, so that&#x27;s it.

Problem

(a) Find the intervals of increase or decrease. …

Problem 29 Hard 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 inforvation from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one.
$h(x)=(x+1)^{5}-5 x-2$

Answer

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

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James Stewart

Essential Calculus Early Transcendentals

Catherine R.

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Caleb E.

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Precalculus Review - Intro

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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

Essential Calculus Early Transcendentals

Catherine R.

Heather Z.

Caleb E.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

James StewartEssential Calculus Early Transcendentals

Catherine R.

Heather Z.

Caleb E.

Samuel H.

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

James Stewart

Essential Calculus Early Transcendentals