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Question 15 ptsConsider the following function f(e) 4x + 1 ifr < 3 kr? ift > 3We would like to find the value of k which makes the function continuous at 3.Fi...

Question

Question 15 ptsConsider the following function f(e) 4x + 1 ifr < 3 kr? ift > 3We would like to find the value of k which makes the function continuous at 3.First of all; the function value {(3)Sclcct ]The limit from the left is lim_-3 - f(r)| Select |The limit from the right is lim, +Jt f(r)=#Select |From this information we can conclude that the value of k which makes f(x) continuous at 3 i5 k= Sclect ]

Question 1 5 pts Consider the following function f(e) 4x + 1 ifr < 3 kr? ift > 3 We would like to find the value of k which makes the function continuous at 3. First of all; the function value {(3) Sclcct ] The limit from the left is lim_-3 - f(r) | Select | The limit from the right is lim, +Jt f(r)= #Select | From this information we can conclude that the value of k which makes f(x) continuous at 3 i5 k= Sclect ]



Answers

Find the value of the constant $k$ that makes the function continuous.
$g(x)=\left\{\begin{array}{ll}{\frac{2 x^{2}-x-15}{x-3}} & {\text { if } x \neq 3} \\ {k x-1} & {\text { if } x=3}\end{array}\right.$

So in order for this function to be continuous G of X x cubed plus K should be equal to uM que X minus five at the values given. So the critical point here is X is three. So let's plug in X here. So three cube that's 27 plus k because that's equal to three K minus five. Let's add five on both sides. That would be 32 plus K is equal to three k continuing on the next pace of 32 plus case three k Subtract k on both sides. Will get 32 is two k. Now divide to on both sides is OK is 16 k should be 16 for their ah for G of X.

Were to find a constant case so that the function is convenience on any interval. The function effects it has given that. Yes Then exorcism 1st 2 3 five When Access greater access vitamin three so it is continues for an interval so we will turn the community as opposed to three. So For the community there is a three. The fo staying Yeah. First type of trouble vehicles to three kids should be defined. Mm and chicken limit Exchange 2 3- of effects that it scares. Yeah, because to limit extent of sleepless five, the promotion of equal. Okay, so it will be close to three K equals two. Bye. When we put the limit and the limit will be exist in their technology goes to NHL. So from here we can find the value of the cable because 25 or three. This is the answer I hope we understood. Thank you.

Okay. So we're gonna be using properties of limits to find um this limit and we're gonna break it down first. So the first property is we have a limit of multiple terms added together. That's equal to the limit of each of those terms separately, then added together. So we're gonna have to limit its K. Goes to two of negative two K cubed. Then plus the limit, His key goes to two five K. And then lastly plus the limitless K goes to two of 9. And so this last limit is just the limit of a constant and we know that any limit of a constant is going to be equal to that constant. So limitless Kgosi two of nine is equal to nine. Um What else I'm going to do is also take out these constants from these two limits. Um Since we're allowed to do that, we can take them out front and multiply them. By the limit in this case would be K cubed and then the other one is going to be five times the limit okay goes to two of just K and then we have plus nine. So the limit is K goes to two of k is going to be equal at two. Um What we can do in this limit is when we have a limit of K cubed we can say that's equal to the limit is K. Goes to two of just K. Then cubed. So this would be equal to negative two times. To limit This key goes to two K. Then cubed and then we already saw that limit of K as key goes to two is equal to two. So it's gonna be five times two plus nine. So here is going to simplify the negative two times two cubed plus five times two is 10 plus nine. two cubed is eight times negative, two is negative, 16 Plus 19 is equal to three.

So we're looking at this limit as K goes to three of the function, we can say F f K here is equal to K. And since this function Um we can FF three here exists if a three here would just be equal to three and three is in our domain, we are continuous and differentiable at that point. And so because of that, what we can do is just directly substitute in three for K here to look at this limit and figure out what it's actually equal to, because ff three is a is a valid point on our function and we are continuous and differentiable at that point. So if we let the limit as K goes to three of this function, f f K is equal to K. Um We just plug in three for K. You can see that this is going to be equal to three.


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