Answer
(a) $f'(a)$ = $\frac{1}{3a^{\frac{2}{3}}}$
(b) This function increases without bound, so the limit does not
exist, and therefore f'(0) does not exist.
(c) f is continuous at x = 0 (root function), so f has a vertical tangent at x = 0
Work Step by Step
(a) Note that we have factored x - a as the difference of two bubes in the third step
$f'(a)$ = $\lim\limits_{x \to a}(\frac{f(x)-f(a)}{x-a})$ = $\lim\limits_{x \to a}(\frac{x^{\frac{1}{3}}-a^{\frac{1}{3}}}{x-a})$ = $\lim\limits_{x \to a}[\frac{x^{\frac{1}{3}}-a^{\frac{1}{3}}}{(x^{\frac{1}{3}}-a^{\frac{1}{3}})(x^{\frac{2}{3}}+x^{\frac{1}{3}}a^{\frac{1}{3}}+a^{\frac{2}{3}})}]$ = $\lim\limits_{x \to a}[\frac{1}{(x^{\frac{2}{3}}+x^{\frac{1}{3}}a^{\frac{1}{3}}+a^{\frac{2}{3}})}]$ = $\frac{1}{3a^{\frac{2}{3}}}$
(b)
$f'(0)$ = $\lim\limits_{h \to 0}(\frac{f(0+h)-f(0)}{h})$ = $\lim\limits_{h \to 0}(\frac{h^{\frac{1}{3}}-0}{h})$ = $\lim\limits_{h \to 0}(\frac{1}{h^{\frac{2}{3}}})$
This function increases without bound, so the limit does not
exist, and therefore f'(0) does not exist.
(c) $\lim\limits_{x \to 0}|f'(x)|$ = $\lim\limits_{x \to 0}(\frac{1}{3x^{\frac{2}{3}}})$ = $\infty$ and f is continuous at x = 0 (root function), so f has a vertical tangent at x = 0.
