Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 42

Determine the values of the constants $a$ and $b$ that make the function $\displaystyle f(x) = \left\{ \begin{array}{cc} \frac{x^2 - 4}{x - 2} & \text{ if } & x < 2 \\ ax^2 - bx + 3 & \text{ if } & 2 < x < 3 \\ 2x - a + b & \text{ if } & x \geq 3 \end{array} \right. \quad$ continuous everywhere.

Based from the definition of continuity,
The function is continuous of at a number if and only if the left and right hand limits of the function at the same number is equal. So,


For $x = 2$


$\begin{equation} \begin{aligned} \lim \limits_{x \to 2^-} \frac{x^2 - 4}{x - 2} &= \lim \limits_{x \to 2^+} ax^2 - bx +3\\ \lim \limits_{x \to 2^-} \frac{(x + 2) \cancel{(x- 2)}}{\cancel {(x - 2)}} &= \lim \limits_{x \to 2^+} ax^2 - bx +3\\ \lim \limits_{x \to 2^-} (x+2) &= \lim \limits_{x \to 2^+} ax^2 - bx +3\\ 2+2 &= a(2)^2 - b(2) + 3\\ 4 &= 4a - 2b + 3\\ 4 - 3 &= 4a - 2b\\ 4a - 2b &= 1 && \Longleftarrow \text{ (Equation 1) } \end{aligned} \end{equation}$


For $x=3$


$\begin{equation} \begin{aligned} \lim \limits_{x \to 3^-} ax^2 - bx + 3 & = \lim \limits_{x \to 3^+} 2x - a +b\\ a(3)^2 - b(3) + 3 & = 2(3) - a + b\\ 9a - 3b + 3 & = 6 - a + b\\ 10a - 4b & = 3 && \Longleftarrow \text{(Equation 2)}\\ \end{aligned} \end{equation}$


Using Equations 1 and 2 to solve for the values of $a$ and $b$ simultaneously.

$\displaystyle a = \frac{1}{2}\\ \displaystyle b = \frac{1}{2}$


Therefore,
The values of $a$ and $b$ that will make the given function continuous everywhere are $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{1}{2}$ respectively.