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Question

For what value of the constant $c$ is the function $f$ continu-
ous on $(-\infty, \infty) ?$
$$f(x)=\left\{\begin{array}{ll}{c x^{2}+2 x} & {	ext { if } x<2} \ {x^{3}-c x} & {	ext { if } x \geqslant 2}\end{array}ight.$$

Agata Pavone · Accepted Answer

Well, let&#x27;s no cities function. Flags equal. It&#x27;s equal to feet. Ex queer plus two eggs you access is less than two. Oh, ex queued My no see eggs if X is greater or equal to and see is just and constant. So let&#x27;s try to work out the value off this constant for the function to be continuous. So we want dysfunction to be continuous in on minus infinitive plus and fill it so everywhere for any really valuable backs. Andi were to work out the value off this constant, which makes the function continuous everywhere. So does let&#x27;s use the definition of continuity. So first the world well, these are just born in no bills, so they are continuous everywhere for any real value. Kovacs. Now the only issue is at two when x equal to this is the only point where the function may up discontinue tive. So in order to get a continuous function at X equal to the usual requirement, af two has to be equal to the limit. Well X, a purchase to of the function. So I fought to the function at X equal to well, it&#x27;s equal. We see when x equal to the function is X cubed minus asiax So it will be X cubed mills to Q minus to see. That&#x27;s the value off I forgot to. We can rewrite it as eight miners to see No, what about the limit? Well, the limit of the function when eggs and porches to from the left is equal to when the function. In this case, the function we have to choose is the 1st 1 see X squared plus two acts. And when X supporters do, this is a continuous function. So what can we write? What we just plug in the valley to So we&#x27;re going where going to on with four C plus four that said the limit when x purchased two from the ride. Well, in this case to limit as going to roar this function execute minus see eggs because X is greater than two on. And what about this payment? Well, this is going to be two to the power off free once more minus two c. So is just eight minus two c, which is exactly I fall too. No, If we want Thies to limits to be the same well, what should we solve? just the question for C plus for the limit from the left has to be equal to the limit from the right a minus two C So that&#x27;s orbit for C means for C class to see Ikhwan a minus four mean six C equal for C equal for over six. We can simplify. We get to over free. Thes is the volume off. See according to which this function is containers everywhere on minus infinity on the interval, minus infinity plus infinity.

Ma. Theresa Alin · Answer

consider the piecewise function F of x equals c X squared plus two X. If x is less than two and f of x equals x cubed minus C X if x is greater than or equal to two. Now, in this problem we are to find the value of C in which these functions continue us on negative infinity to infinity. Notes that this function is already concerned was over the interval negative infinity to two. And over the interval truth infinity. This is because the function is defined in ancient herbal as a polynomial. And so we need only show that the function is continuous at X equals two. That is we need to show that the limit as X approaches to of dysfunction exists. Now this exists if The one sided limit limit as X approaches, let&#x27;s say two from the left of this function is equal to the limit As extra purchased two from the right of dysfunction. Now, the limit as X approaches to you from the left of this function, this is just limit As X approaches to from the left of, we will use the um function C X squared plus two X. Since X approaches to Two from the left, that means X must be less than two. So we do see X squared plus two X. And then from here we want to evaluate the function at two and so we get C times two squared plus two times to that four. C Plus four. And then the limit of the function as X approaches to from the right, this is equal to the limit as X approaches to from the right of the function X cubed minus c. X. We&#x27;re using this because the Values of x here are those greater than two. And so we have evaluating it to we get to to the third power- C, Times two. That&#x27;s just 8 -2 C. And because you want this function to be continuous then they&#x27;re forced to yeah, make the one sided limits equal means we right or C Plus for this should be equal to 8 -2 c. and solving for C, we have four C plus to see this is equal to eight minus four. And so six. See this is equal to four or that C is equal to 4/6 or 2/3. So this is the value of C in which the function is continue was over the interval negative infinity to infinity.

Paul Teng · Answer

looking for this from that, I want you to differentiate the function 1/8 times. Expert close. Moreover, be times experts seem this is a polynomial consists off some power functions like such as X squared. So we&#x27;ll be using a power role and the power role states if we want to differentiate a power function, we just have to bring down the excrement and multiplied by accident and minus one. So we&#x27;ll be using that. We could take the derivative of f of X becomes after prime of X equals two or one over. A is a constant coefficient in front of some variables, so we can leave that alone at the beginning of the pregnancies. And now we have differentiate this. So the derivative of X square, using the power room is to accept that the bring down the to and subtract one from the Excellent. Now the derivative of one Overbey time ex is just one over being. Because whenever b is a cult coefficient constant, so we leave that alone like probably left one over. A home and derivative of X is just one, and finally the derivative of C. It is just a constant pussy. It&#x27;s just a constant. So if you differentiate that it becomes zero, this become zero. So the answer is just 1/8 times two x plus one over beat.

Paul Teng · Answer

So this problem wants to take the derivative of the function eight times extra cubed plus B X squared plus c expose D or A, B and C and D are all constants. So this is a polynomial with a bunch of power functions. So we&#x27;ll be using the power rule. Take the derivative its function and the power rule says that we want take a derivative of power function will have to bring down the exponent and multiplied by X city and minus one. So we&#x27;ll be using this a lot. And the constants A, B, C, NT are all coefficient constants. Will A, B and C your crawfish and constant soul. Do you believe those things alone as we solve the problem? So a since I was just a constant in front of X cubed we&#x27;ll leave that alone so we just don&#x27;t differentiate, execute and using the power rule that becomes three x square. Next is B X squared. You know that B is a coefficient constant, so we&#x27;ll leave that alone. And the drill of X squared by the power role is two times X since two minus one is just one. So just accept one power, but we can just think about it. Just acts and see time on C Times X on the derivative of X is just one. So we just getting see as a constant and finally the derivative of D, which is a constant without an X variable, so that just become zero. Simplifying this, we&#x27;ll get a form of X equals two a. Three a x squared was to be times X plus c, and this is your final answer.

Carter Rogers · Answer

here. We&#x27;re looking at a piece wise function F of X equals three X plus b, where B is some number if X is less than or equal to two and X minus. Two of X is greater than two that we&#x27;re being asked to evaluate the limit at X Purchase two of FX. But in order to do that, we need a value of B that allows us to occur. We know that in order for this limit to exist, the limit from both sides has to be equivalent. So the limit is X purchase to from the negative side of F of X has to equal the limit has exit purchase to from the right, So we&#x27;ll go ahead and do some algebra to verify that. So the limit as expertise to from the negative side that gives us three X Plus B and the limit as exit purchase to from the right side. It gives us X minus two and again, we&#x27;re taking the limit as experts to from the left and from the right, so we&#x27;ll go ahead and plug in the X value so we get three times two plus B playing in two for X here we get to minus two. So over here we get Sabih over here we get six plus B equals zero. In other words, be equals negative six. So in order for this limit to exist, B has to be negative six and then we&#x27;ll go ahead and evaluate The limit has expertise to of FX. So we get the limit as experts is to from the left side. Now, using F of X has three X minus six. This gives us three times to minus six, which is zero. And we already evaluated the limit over here, which is zero. So the limit as f of X, a purchase two overall is going to be zero.

Jason Orozco · Answer

we want to know the value of the constant A for which this peace WAAS function is continuous at X equals negative one. So since the function equals a at X equals negative one, um, we really just need to set a equal to the limit of the other piece of the function in order to satisfy the rules for continuity. So, um, in order that we just need to figure out what that limit its so, uh, running that down, we have limit as X goes to negative one of X squared plus three x plus two over X plus one. And, uh, we can&#x27;t directly substitute because we get a zero the denominator. But it&#x27;s a Russian polynomial. So if we can, uh, we could potentially remove a discount, continuity, um, and go from there. So in order to do that, we&#x27;re gonna factor the numerator limit of X as X goes to negative one of X plus two times X plus one because two plus one is three and two times one is two all over X plus one. So the X plus one factors cancel out and we can just plug negative one into exposed to So we get negative one plus two, which is one. So the limit of the the variable piece of the function as X goes to negative one is one which means that, um if we if we set a equal toe one, then the limit equals the value of the function. So a equals one is our solution.

Problem

Find the values of $a$ and $b$ that make $f$ cont…

Problem 33 Medium Difficulty

For what value of the constant $c$ is the function $f$ continu-
ous on $(-\infty, \infty) ?$
$$f(x)=\left\{\begin{array}{ll}{c x^{2}+2 x} & {\text { if } x<2} \\ {x^{3}-c x} & {\text { if } x \geqslant 2}\end{array}\right.$$

Answer

$$
c=\frac{2}{3}
$$

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James Stewart

Essential Calculus Early Transcendentals

Anna Marie V.

Kayleah T.

Caleb E.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$$c=\frac{2}{3}$$

Essential Calculus Early Transcendentals

Anna Marie V.

Kayleah T.

Caleb E.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

James StewartEssential Calculus Early Transcendentals

Anna Marie V.

Kayleah T.

Caleb E.

Samuel H.

$$
c=\frac{2}{3}
$$

James Stewart

Essential Calculus Early Transcendentals