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Given the function;4x2 _ 4x - 4 ifx < a flx) = 6x2 +x _ 2 ifx > aDetermine the the value or values of that will make the function continuous on 3,0)...

Question

Given the function;4x2 _ 4x - 4 ifx < a flx) = 6x2 +x _ 2 ifx > aDetermine the the value or values of that will make the function continuous on 3,0)

Given the function; 4x2 _ 4x - 4 ifx < a flx) = 6x2 +x _ 2 ifx > a Determine the the value or values of that will make the function continuous on 3,0)



Answers

The function $f(x)=\frac{x^{2}-4 x+3}{x^{2}-1}$ is continuous over the interval $[0,3]$

Hello. And they're giving function fx This. Is it close to experts for up on texas car minus X. Okay. And you can see that we can try to like this X plus four upon X and two X minus fun. This is a rational function and we know that this function is continues for every X. Except except that did not accept the value of X. The value of X. Very nanometers and your cost to zero and you can see the denominator is X into X. Man advantage. Request to Jerusalem from here. X across zero and one for these two values at As close to zero and 1. The funds are not continuous. Yes, not continuous because at this point the functions are defined. Had these values funds are not define. So this is the answer. I hope you are not sure. Thank you.

So we have F of X is equal to X to the fourth minus two x to the third, divided by X minus two. And if we look at this function on the interval negative 3 to 3, we know it's not entirely continuous because at X equals two, the denominator is equal to zero and therefore undefined because you can't divide by zero now because, um, the it's not always continuous on negative 3 to 3, it does not satisfy the intermediate value serum sad. We show that we were to graph this function on and on a coordinate plain. It would look like this. It would look like a regular, um, cubic function. Um, now that's very it would look like a regular cubic function. But where, UM, at X equals two, there would be a removable disk continuity because we already said At X equals two, it's undefined. So what is the intermediate value? Theorem says? It says that for every, um, for every if we look at f of negative three, I will say is right there and f of positive three, which will say is right there. Then we know that for any why value K wherever we played some. Okay, um, there is a an X value within the interval. Negative 3 to 3. Such that, um f of that X value is equal to K. Basically, what this is saying is, if we draw a horizontal line, y equals k anywhere between negative three and three, then at some point f of X has to intercept y equals k between negative three and free. And we know this doesn't work, because if we were to draw a vertical line are a horizontal line right here that passes through the, um removable disk Continuity at X equals two. Then there is no value on f of X, such that f of X intercepts y equals K because the only value that would possibly work is at X equals two, which we already know is undefined. So So that means it does not satisfy the intermediate value theorem because there is some value k that is not intersected by f of X between negative three and three

So we have the piece wise function ffx so f of X equals B X plus four when X is less than or equal to three and f of X is equal Toby X squared minus two when X is greater than three. So we want to find value a value of be such that, um such that, um f of X is continuous at X equals three. So to do that, we first have to show what f of three is. So it says when um X is equal to three. We use the top equation. So that's, um three B plus four. Now, to find the we have to determine the limit whether it exists. So the limit of f of X as X approaches three from the left. First of all, we're looking for values less of X. Lesson three. So again, we use the top equation. So that's just again we can plug in threes. This again becomes three b plus four from the right. We used the bottom equation because those air X values greater than three. So again we substitute in three for X of three. Squared is nine B minus two now in order for the limit to exist. The right and left hand limits have to be equal. So if we set up that equation three B plus four equals nine B minus two, we can find, um, what value of B Let's them be equal. So if we subtract herbal sides by three b, we get four equals six B minus two. Add to double size. We get six equals six b. So be is clearly one. And we can check that when, um if B is one than the limit becomes, um, three times one plus four, which is equal to nine times one minus two. Three times one plus 47 equals nine minus two, which is seven. So the limit exists when B is equal to one. Now let's go back to F of three and substitute in B is equal toe one. So we get three times of one plus four again, that's equal to seven. Um, so because F of three exist and the limit exists and they are equal so one another, then it means that it's continuous. F of X is continuous at X equals three when be is equal to one

Bring it here and 10. We only really need to look at the denominator, since they're both polynomial. And since our denominator is X squared minus one, that implies that it'll equals zero and X is equal to plus or minus one, just due to the difference of squares. If you remember, this would simplify two x plus one times X minus one and saying that equals zero because of the zero, property has three plus or minus one. And that means the interval for which this will be continuous will go from negative infinity to negative one union with from negative one to positive one union with from positive one to infinity. So just the two continuity errors at negative one and positive one. Thank you very much.


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