How to prove a function is not differentiable

Question

Given:

$$f(x) = \begin{cases} x^3 &&& x<2 \\x+6 && &x \geq 2 \end{cases} $$

I need to prove that $f(x)$ is not differentiable at $x=2$, what should I do? $$\lim_{x \to 2^+} \frac{f(x) -f(2)}{x-2} =\lim_{x \to 2^+} \frac{x+6-8}{x-2} = 1 $$ $$\lim_{x \to 2^-} \frac{f(x) -f(2)}{x-2} =\lim_{x \to 2^+} \frac{x^3 -8}{x-2} = 6 $$ Is taking limits for the two functions and showing that those two are not equal correct?

Answer 1

You can also do the alternative, which is take the symbolic derivative of $f(x)$. The result is:

\begin{align} f'(x) = 3x^2 & , 1 \\ f'(2) = 12 & , 1 \end{align} and $12 \neq 1$.

Answer 2

Use the definition of continuity: $\lim_{x \to x_0} f(x) = f(x_0)$. Clearly $\lim_{x \to 2^{-}}3 x^2 = 12 \neq 1 = f(2)$.