a-find-the-intervals-of-increase-or-decrease-b-find-the-local-maximum-and-minimum-values-c-find-t-21

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.
$C(x)=x^{1 / 3}(x+4)$

Jack Chen · Accepted Answer

for this question, we need toe right out the narratives for see at the beginning and the second with the narrative. So for Papuan, we want to find the increasing and be Christine Herbal. So he says, See prime to be zero. Also, we have X equals of minus one, but we also need to consider um, the frost majority. It does not exist. So when X equals zero, that&#x27;s the first solidarity. If that&#x27;s not exist, that means we need to consider interval from minus infinity to minus one and from minus 20 0 and from zero to infinity over the first interval. See prime this positive. So sits up. See price and active. The function is decreasing over the second hurdle. See promise positive. The function is increasing over the last it Roxy Prime Minister also positive the functions increasing that means for palatable. We have a local lenamon that x the coast of minus one with values see off minus or in Khost of miners green. And for posse, we said to say, for the purity of to be zero. So we have XY closer to again. We need to consider X equals zero, because at this point on the Saco majority of does not exist. So we have to wear three sub intervals from minus infinity to zero from 0 to 2 in the front. Tool to infinity over the first interrupt. See, double prime is positive, so it functions can keep up over the second. Horrible. See, double prints, connective. The functions can keep down over the last interval. Um, the second of narrative is positive. The of the functions come cave up. So in fact, with two inflection points between connection points head, um, X equals zero in the XY quest to Based on this information, we can schedule a graph for faith. It&#x27;s the first relabeled some the critical point X equals one minus one and X equals zero. And we have an inflection point at zero and two. Okay, so the function is decreasing first, and they it has a local minima taxi Costa minus one on day increasing. So the function we passed this to your point listed minus 40 and the so basically, we have something like this. So when exist, lesson zero, the function is actually come cave, um, his conclave off based on for a second. They&#x27;re from zero to the FARC is concave down something like it is. And we passed the point to function the country pop again. So the phone function looks actually, So we have to inflection points. So we must change the contribute e. And then we have a local minimum, which is also a global meaning that we can see that and we have two x intercept.

Clarissa Noh · Answer

he is clear. So when you raid here, So for part A, we&#x27;re gonna find the derivative of C of X, and we get for X plus four over three X to the 2/3 so we&#x27;re gonna find when it&#x27;s increasing and decreasing. It&#x27;s increasing when the derivative is bigger than zero and decreasing when a smaller than zero. So we see that it is increasing when exes between one negative one and infinity and increasing when it&#x27;s between decreasing. When it&#x27;s between negative infinity and negative one For part B, we see a local minimum at sea of negative one, which is equal to negative three. Since that changes from decreasing to increasing report. See, we&#x27;re finding the second derivative. Let me get for X minus two over nine x to the 5/3 and this is where the second derivative equal to thorough. So we get X is equal to two, and when X is equal to zero, it&#x27;s undefined. So we take the intervals Negative infinity comma zero. We choose C double derivative of negative one and this is bigger than zero, which means it&#x27;s calm. Keep up between the interval zero and two. We choose one just less than zero in between two and infinity, you choose three, which is bigger than zero. So the changes con cavity at zero and two. So there are inflection points zero comma zero into comma 7.56 When we plug it into the orginal equation Part D we just draw a graph and I love something like this

Madi Sousa · Answer

and this problem, we are learning how derivatives affect the shape of a graph. More specifically, this problem is asking us how we can use the derivative to deduce important information about the function, like where the function is increasing or decreasing three intervals of con cavity. And it&#x27;s critical numbers. And this is going to become important later on calculus when you start to learn curve sketching, so getting a great understanding of this now is going to be very helpful for you. Sorry. Pardon me for part A were given the function F of X equals in one half X to the fourth minus four X squared plus three. So we need to find the critical numbers and the intervals of con cavity. So the first thing that we&#x27;re going to do is take the derivative F prime of X equals two X cubed minus eight x We could do some. Factoring will get two x times X minus two times X plus two. We&#x27;re going to set that equal to zero to get the critical numbers. Then we&#x27;ll solve for X so x zero X s two and excess negative too. So then what? We&#x27;re going to Dio is we&#x27;re going to create a table. You don&#x27;t have to make a table, but just keep track of the intervals. We&#x27;re going to make sure that we have the intervals using the critical numbers and then determine if our function would be increasing or decreasing in that interval. So the first interval is negative. Infinity too negative too. And our function is decreasing there. And if you don&#x27;t know how I&#x27;m getting that, all you have to do is take a number within that interval, plug it into the function and see if you get a positive or a negative. For the second interval, we&#x27;d have negative 2 to 0 and we see our function is increasing there. Our third interval interval. Pardon me is from 0 to 2 and our function is decreasing there. And then finally, our last interval is to to infinity and our function is increasing there. So this is just another way to right. We found on the table at the vex increasing when X is an interval. Negative 20 and two comma infinity and f of X is decreasing on the interval when X is in thes two intervals negative and they needed to negative negative two or two or pardon me, you 02 and then for party were told. What are the maxes and men&#x27;s of this function? What we have the critical numbers. All we have to do now is plugged them back into our original function. So f of negative two is negative. Five f of two was negative. Five and F of zero is three. So clearly we have a minimum, a negative two and two and a maximum at zero for part C, we&#x27;re told. Let&#x27;s check the con cavity. So when you hear the word con cavity your mind, you immediately go to second derivative. We need the second derivative test to determine the intervals of con Cavity. So F double prime of X is six x squared minus eight. And now this one&#x27;s a little bit more confusing than before. We have to do some clever factoring. So this is the same thing as saying six times X squared, minus 4/3. And then we want critical numbers, Remember? So we have to factor this further. We&#x27;re going to get six times X minus two over the square root of three times X plus two over the square root of three. So then we have our critical numbers so we can determine con cavity. Our function is con cave up on the interval. Negative infinity negative to over screwed of three and two over. Screwed of three common infinity And then we see a con cave down behavior on the interval negative to over the square root of 3 to 2 over the square root of three. So what does this mean? We change from going from concave up to con cave down So we see an inflection point when x is negative two over the skirt of three and when X is two over the square root of three. And now finally the problem asks us, Well, what&#x27;s going to happen with the graph? Can we check our information from the graph? This is the graph of the function and you can clearly see that our information that we found using just differential calculus matches the function. We see our minimums where we found them, our maximum where we found them. Our con cavity matches as well. So I hope this problem helped to understand a little bit more about how the derivative, um, affect the shape of the graph more specifically, how we can use the derivative to deduce important information from a function such as the intervals of increase in decrease and it&#x27;s common cavity.

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.

Jack Chen · Answer

for this problem its first right out euro tips the front both love for so that he right here on the second narrative off f So pat A We set if prime Michels to zero So we have X equals to pass minus two. That means we need to consider the recipient. How goes negative infinity to connective to connective 2 to 2 and the two to infinity First on the first interval. If prime is positive safety increasing on the second of life promise negative So it&#x27;s decreasing I&#x27;m allows him however, if crimes positive again city increasing and from part of a we know that when X equals two minus two So there&#x27;s a local, uh, maximum with value f to you crossed very king and at X equals to two is a local minimum with the value of two equals two minus 14 and the pop See, we said that if double Primakov zero we only have one solution X equals zero So we have to sub intervals from negative infinity to zero and the from zero to infinity on the first sub in trouble If the prime is negative so to come cave down world and Ah, on the second row. If that about Prime miss positive. So it&#x27;s conch it upward. So we have the inflection point at X equals zero, and we are ready to graft such function on the coordinates. So first we can label out some, um, create a point and the inflection points. So, uh, XY caused a minor stories. A local maximum exit question, too, is a local Milliman. And the inflection point is, every issue should be X equals 20 Um, someone X equals zero. The functioning crystal too. Yeah. Label the inflection point here. Okay, on Now we already the graph. So from negative infinity connective to the function is increasing the U eventually and it&#x27;s concave down. So it looks like this. And then when he passed through this local maximum, it&#x27;s d crazy and also calm cave down. Now we passed this inflection point You changes that can cavity to become cave up And we passed this commitment you see increasing again. So on the graph Looks like it&#x27;s This is an X Texas Why exists the three Why existed that this is a

Kian Manafi · Answer

So in part a we&#x27;re going to find where our function F of X is increasing and decreasing. So I&#x27;ve just written down the derivative already and I just use the power rule to find this. And so we&#x27;re just going to figure out where are derivative is positive and where it&#x27;s negative. So the easiest way to do this I think is to find where it&#x27;s actually equal to zero first. Since then we can know what intervals to look at to see how our the sign of our derivative F. Prime of X. And so I&#x27;m just solving for X. Here we get four is equal to X squared, so X is equal to plus or minus two. So a plus or minus two. We have f prime of X equaling zero. And now if we look at the intervals X being less than negative two, X being between negative two and two, and X being greater than X being greater than two. Sorry. And we look at the sign of our derivative, we&#x27;ll know where we&#x27;re increasing and decreasing. So if we plug in an X value less than negative two, let&#x27;s just say negative three. We get nine times three minus 12 which is positive. So are derivative. F prime of X is greater than zero. We plug in a value between negative two and two. Let&#x27;s just say zero. We get zero minus 12 which is negative. So are derivative F prime of X is less than zero. And then if we plug in the value of X greater than two, say three, we get nine times three again, 27 minus 12 is positive. So the intervals in which were increasing, we&#x27;re going to be the first and last intervals. So from negative infinity to negative two and from two to infinity. And then we&#x27;re decreasing in this other interval from negative to 22 Now, if we go to part B, we&#x27;re going to do is find local maximum and minimum values. And we already found the zeros of our derivative up here at plus and minus two. So we&#x27;re just going to look at actually, we&#x27;re just going to look at the information we already found. So you want to look at values of X less than negative two. We can see that we&#x27;re going positive and the values after negative to our negative. So we&#x27;re going from increasing slope to decreasing slope at X is equal to negative two, which means we have a maximum value. So we have a local max at X is equal to negative two. And if we look at the values before too, but greater than negative two were negative and then greater than two were positive. So we&#x27;re going from negative values to positive values are decreasing slope to increasing slope, which means it is a local minimum. And now for part, see what we&#x27;re gonna do is find the intervals in which our function f of X is concave up in concave down. And then we&#x27;re also going to find the intervals are the points of inflection. So I&#x27;ve written down this first derivative because we&#x27;re going to use it to find our second derivative, which then we can use to figure out con cavity. So I&#x27;m just gonna use the power rule again, find the second derivative, bring down the exponents gets six X. And this 12 goes to zero. And now this one we just want to look at where this is negative and where it&#x27;s positive. Well you can see that&#x27;s gonna be negative whenever X is negative, it&#x27;s gonna be positive whenever X is positive. So we&#x27;re concave up from zero to infinity. Since we need positive values are second derivative to be concave up, Merkel cave down from negative infinity to zero. And then we also have an inflection point at zero, since that&#x27;s where our second derivative is equal to zero. And we know we&#x27;re going from negative to positive. So we&#x27;re actually crossing that X axis at X is equal to zero, so inflection point at X is equal to zero. And if we plug that back into our original equation this is actually equal to two. So we have the inflection point is at this 0.0.0 comma two. And now for part D, what we&#x27;re gonna do is use all this information that we found and actually try and sketch a graph of of our functions. So we know that F zero is two, so zero comma two. And that&#x27;s an inflection point, we know that F of zero is, oh never mind. We know that. We know the F. Of zero is too. So I plugged that point in here and now we want to look at, well we&#x27;re concave up before we&#x27;re sorry we&#x27;re concave down when we&#x27;re negative. So I&#x27;m just gonna put concave down, concave up. So I remember and then we have a local max at X is equal to negative two. A local men X is equal to two. So um what we&#x27;re gonna do now is I&#x27;m actually going to just plug in F of two into our equation which was X the third minus 12 X two to the third minus 12 times two plus two. This is equal to eight minus 24 plus two. Just equal to negative 14. So we have this point at to negative 14. I&#x27;m just gonna put down here I&#x27;m not going to get super specific and now I&#x27;m gonna figure out f of negative two. So the only difference here is we&#x27;re gonna have plus 24 instead of minus and this is going to be eight plus 24 32 34. So you have a point F is equal to negative two 34. Just stay up here and now we can look at and this is at negative two and we know that this is a maximum value. This is a minimum value. Mhm. And now we can look at we&#x27;re increasing from negative infinity to negative two and we are concave down right, so we need to be increasing in concave down and then we&#x27;re decreasing in concave actually instead of so we&#x27;re increasing in concave down. And then at this point here too we start to be concave up since that&#x27;s an inflection point and we are now still decreasing until we get to this point at two comma negative 14 and then we start increasing again. So we need to go all the way down into this point to and we can start increasing again and actually shouldn&#x27;t make it look like we start to go concave down again since we don&#x27;t have another inflection point. So this is somewhat of what our graph should look like. Again, we have these local max is that x is equal to negative two. Local men addicts is equal to two. We&#x27;re decreasing from negative to to to or increasing from negative infinity to negative two and from two to infinity. And we&#x27;re concave up where we need to be re concave down where we need to be. We have this inflection point at zero comma two. So yes, this is indeed a good rough sketch of our graph.

Problem

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

Problem 33 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.
$C(x)=x^{1 / 3}(x+4)$

Answer

A. $c(x)$ is increasing when $x \in(-1, \infty)$
$c(x)$ is decreasing when $x \in(-\infty,-1)$
B. Local minimum $=C(-1)=-3$
C. $\mathrm{CU} :(-\infty, 0),(2, \infty) \quad \mathrm{CD} :(0,2) \quad$ inflection points $(0,0),(2,6 \sqrt[3]{2})$
D. SEE GRAPH

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James Stewart

Essential Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

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A. $c(x)$ is increasing when $x \in(-1, \infty)$$c(x)$ is decreasing when $x \in(-\infty,-1)$B. Local minimum $=C(-1)=-3$C. $\mathrm{CU} :(-\infty, 0),(2, \infty) \quad \mathrm{CD} :(0,2) \quad$ inflection points $(0,0),(2,6 \sqrt[3]{2})$D. SEE GRAPH

Essential Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

James StewartEssential Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Kristen K.

A. $c(x)$ is increasing when $x \in(-1, \infty)$
$c(x)$ is decreasing when $x \in(-\infty,-1)$
B. Local minimum $=C(-1)=-3$
C. $\mathrm{CU} :(-\infty, 0),(2, \infty) \quad \mathrm{CD} :(0,2) \quad$ inflection points $(0,0),(2,6 \sqrt[3]{2})$
D. SEE GRAPH

James Stewart

Essential Calculus Early Transcendentals