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Find the values of $ a $ and $ b $ that make $ f $ continuous everywhere.

$ f(x) = \left\{

\begin{array}{ll}

\dfrac{x^2 - 4}{x - 2} & \mbox{if $ x < 2 $}\\

ax^2 - bx + 3 & \mbox{if $ 2 \le x < 3 $} \\

2x - a + b & \mbox{if $ x \ge 3 $}

\end{array} \right.$

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Campbell University

Harvey Mudd College

Baylor University

Idaho State University

and this problem we need to find the values of A and B. That makes this piecewise function continuous everywhere. Now to do that we have to take the limit of the function and x values where The function is described in two ways. And so this will be at X equals two and at x equals three. We're doing this because you want to make sure that the function will take on the same value as X approaches this number, either from the left or from the right. Now, the limit of the function as X approaches to from the left is equal to limit as X approaches to from the left of you have X squared minus four over x minus two, which is the same as the limit as X approaches to from the left of we have X -2 times x plus two Over X -2. this will reduce to express to and so evaluating at two, we get two Plus 2 which is four. While the limit as X approaches to from the right of f of x, this is equal to limit as X approaches to from the right, ah A x squared minus b X plus three. And so evaluating it to we get eight times 2 squared minus b times two plus three. This is just four A minus to b plus three. Now, since we want the function to be continuous then we want the limit as X approaches to of this function to exist. That means the one sided limits must be equal has. Since the one sided limits are equal then we have four a minus to B plus three. This is equal to four. Or if you simplify this, we get For a minus to be, this is equal to one. So this is our first equation for the limit of the function at ah X equals three. We have limit as X approaches three from the left of F of X. This is just limit as X approaches three from the left of A, X squared minus B. X plus three. And evaluating A three, we get eight times three squared minus B times three plus three. This is just nine A -3 B Plus three. For the limit as X approaches three from the right, we have limit As X approaches three from the right of F of X. That's just Limit as X approaches three from the right of two, X -A Plus B. That's just two times three minus A plus B. That's just 6 -5. And because you want to Make the function continuous at three, then we want the limit to exist As X approaches three and so nine A -3 B Plus three. This is equal to six -A Plus B. Simplifying this, we get 10 a -4 B. This is equal to three and this is our equation too. And so we have the system with equations For a minus to be, this is equal to one And we have 10 a -4 b. This is equal to three. Now multiply equation one by -2 and adding it to equation to, we have Have negative eight a. And then this is plus four B -2. So adding this we will get to A and this is equal to one or A is equal to 1/2. Now if a is 1/2, then substituting this either from for equation two or Question one, we get four times 1 half Minour to be, this is equal to one. You get to minus to be that's one or -2 B equals negative one and we get be equal to one half. And so these are the values of A. And B.