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For what value of the constant € is the function f continuous on (~, 0)?cx+7 for X<2 f(r)= cr _3 for x>2...

Question

For what value of the constant € is the function f continuous on (~, 0)?cx+7 for X<2 f(r)= cr _3 for x>2

For what value of the constant € is the function f continuous on (~, 0)? cx+7 for X<2 f(r)= cr _3 for x>2


Answers

Find the constant $a$, or the constants $a$ and $b$, such that the function is continuous on the entire real number line. $$f(x)=\left\{\begin{array}{ll} 3 x^{2}, & x \geq 1 \\ a x-4, & x<1 \end{array}\right.$$

For this problem we were given a piecewise function and has to find the values of A. And B. Which are both constants such that it's going to be continuous for the entire number line. So first we need to see the values where the functions change. And two values we have our negative one and three because from a negative one we go from two to A X plus B. And three we go from X plus B. Two negative two. So what we need to do is evaluate the functions at these values. Well since the first we have to for negative one, well choose a constant. So it's not going to be affected by X values. So we simply equal to equals to A X plus B. And for the first one it's gonna be X negative one. So two is going to equal to negative A. Plus B. And that's what we get left with. And then on the other side we need to do it at three. So negative to again is a constant. So the X values changing aren't gonna affect uh the negative too. So -2 needs to equal to A. X plus B. When X is negative or sorry when X is equal to three. So negative two is going to equal to three X. Sorry three a. rather plus B. And then what we can do is at the two together. So let's rewrite them both right next to each other. You plus B. That is A plus B. Negative two equals three A. Plus B. We add them together and we get zero is equal to two A. Plus two P. And that's gonna be to -2 a. is equal to negative. To be sorry, positive to me or negative A. Is equal to be. Now since we have this we can plug in beef up here so we're going to get to is equal to B plus B, or to be. And so too are sorry. And so B is going to equal to one. And now we can plug that in over here. So negative two is equal to three. A plus one subtract from both sides. Negative three is equal to three. A Or a is equal to -1. And so our two values are A. Is equal to negative one, and B is equal to one.

Hello. So here we have the fx is either equal to two. If X is less than or equal to negative one or equal to X plus B. If we have X between negative one and three or is equal to negative two If X is greater than or equal to three. So um for efforts to be continuous on the entire rail line, if it's continuous on X is equal to negative one and 0 to 3. So we have to have the limit as X approaches negative one, our exports is one from the left to be equal to f of negative one. The limit as X approaches three from the left um and from the right most must be equal to every three. And the limit as X approaches one from the left and the right must be equal to um to f of negative one. So now for continue continuity at X equals negative one. We must have the limit as X approaches native one from the right. Um to be equal uh after backs to be equal to two which implies that the limit As X approaches -1 from the right of a X plus B is going to be equal to two which implies that negative A plus B is equal to two by direct substitution. Um And for X is equal to three. We have to limit as exporters three from the right um is going to be equal to um negative to what's going to be equal to f of three and for continuity at X equals three. We must have the limit exposures three from the left Um after backs to be negative too, which implies that the limit as X approaches three from the left of A, X plus B is going to be equal to negative two, which implies that three A plus B right, three A plus B is equal to negative two's by direct substitution. And from there we get that four A is equal to negative four. So we did that for A Equals -4, which gives us that a Is equal to -1. And um from there we can put that value in and would that be is equal to one? Which implies that A is negative one and B is one.

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'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'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's just 8 -2 C. And because you want this function to be continuous then they'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.

But this problem, we are given the function F of X equals two X plus nine. X rays are negative one where X is less than or equal to three and negative four X plus C. Where X is greater than three. We are asked to find the value of the constant C. In this case that makes the function continuous. So what we need is that two times 3 Plus nine, divided by three, must equal -4 times three plus C. Where we can see that to temperature is going to be 6. 9/3 is gonna give us three, negative four times three is going to give us negative 12. We have the plus C. We can rearrange this very easily. Then for C. Uh six plus three would be nine, then we add on 12. So we'll have that C must equal 21.


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