Find values of constants so that the derivative of a function exists

Consider the function $f$ defined by

$f(x) = \begin{cases} x^2 & \text{if } x \leq c, \\ ax+b & \text{if } x > c. \end{cases}$

Find values for the constants $a$ and $b$ such that the derivative $f'(c)$ exists.

We know that the derivative $f'(c)$ exists if and only if

$\lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$

exists. Furthermore, this limit exists if and only if the one-sided limits both exist and are equal:

$\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}.$

So, plugging in the formula for $f$ (which is $ax + b$ if we approach $c$ from the right, and is $x^2$ if we approach from the left, and noting that $f(c) = c^2$ from the definition of $f$ ) we have,

$\begin{align*} &&\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} &= \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} \\ \implies && \lim_{h \to 0^+} \frac{a(c+h) + b - c^2}{h} &= \lim_{h \to 0^-} \frac{(c+h)^2 - c^2}{h} \\ \implies && \lim_{h \to 0^+} \frac{ac + b - c^2}{h} + a &= 2c. \end{align*}$

For the limit on the left to exist we must have $ac + b - c^2 = 0$ (otherwise the limit will diverge as $h \to 0$ ). Furthermore, this limit must be 0 since $ac +b - c^2$ is a constant (and the limit of $0/h$ as $h \to 0$ is 0). Therefore, we have $a = 2c$ , and we have the equation