Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 96

Find the values of $m$ and $b$ that make
$
f(x) = \left\{
\begin{array}{c}
x^2 & \text{if} & x \leq 2\\
mx+b & \text{if} & x > 2
\end{array}\right.
$


In order for the function to be differentiable everywhere, the function should be continuous
on every values of $x$, that is, the left and right hand limits as $x$ approaches 2 should be equal.

$f'_- (2) = f'_+ (2)$

$
\begin{equation}
\begin{aligned}
\lim\limits_{h \to 0^-} \frac{f(x+h)-f(x)}{h} &=
\lim\limits_{h \to 0^+} \frac{f(x+h)-f(x)}{h}\\
\lim\limits_{h \to 0^-} \frac{(x+h)^2 -(x)^2}{h} &=
\lim\limits_{h \to 0^+} \frac{m(x+h)+b-[mx+b]}{h}\\
\lim\limits_{h \to 0^-} \frac{\cancel{x^2}+2xh+h^2-\cancel{x^2}}{h} &=
\lim\limits_{h \to 0^+} \frac{\cancel{mx}+mh+\cancel{b}-\cancel{mx}-\cancel{b}}{h}\\
\lim\limits_{h \to 0^-} \frac{\cancel{h}(2x+h)}{\cancel{h}} &=
\lim\limits_{h \to 0^+} \frac{m\cancel{h}}{\cancel{h}}\\
\lim\limits_{h \to 0^-} (2x+h) &=
\lim\limits_{h \to 0^+} m\\
2x+0 &= m
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
m & = 2x \quad ; \quad x = 2\\
m &= 2(2)\\
m &= 4
\end{aligned}
\end{equation}
$


Solving for $b$,
$x^2 = mx+b$


$
\begin{equation}
\begin{aligned}
(2)^2 &= 4(2) +b\\
b &= 4-8\\
b &= -4
\end{aligned}
\end{equation}
$

Therefore, the values of $m$ and $b$ that will make $f(x)$ differentiable every where are 4 and -4
respectively.