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.