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.