Answer
(a) $f'(a) = \frac{1}{3a^{2/3}}$
(b) $f'(0)$ does not exist.
(c) $y = \sqrt[3] x~~$ has a vertical tangent line at $(0,0)$
Work Step by Step
(a) $f(x) = \sqrt[3] x$
We can find $f'(a)$:
$\lim\limits_{x \to a}\frac{f(x)-f(a)}{x-a}$
$= \lim\limits_{x \to a}\frac{\sqrt[3] x-\sqrt[3] a}{x-a}$
$= \lim\limits_{x \to a}\frac{\sqrt[3] x-\sqrt[3] a}{(x^{1/3})^3-(a^{1/3})^3}$
$= \lim\limits_{x \to a}\frac{\sqrt[3] x-\sqrt[3] a}{(x^{1/3}-a^{1/3})(x^{2/3}+x^{1/3}a^{1/3}+a^{2/3})}$
$= \lim\limits_{x \to a}\frac{1}{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}}$
$= \frac{1}{a^{2/3}+a^{1/3}a^{1/3}+a^{2/3}}$
$= \frac{1}{3a^{2/3}}$
(b) We can try to find $f'(0)$:
$f'(a) = \frac{1}{3a^{2/3}}$
$f'(0) = \frac{1}{3(0)^{2/3}}$
This expression does not exist when $a = 0$
Therefore, $f'(0)$ does not exist.
(c) Note that $f(0) = 0$
The slope of a tangent line at a certain point is equal to the slope of the graph at that point. The slope of a graph at point $a$ is $f'(a)$
We can find $f'(a)$ as $a$ approaches 0:
$\lim\limits_{a \to 0}f'(a) = \lim\limits_{a \to 0}\frac{1}{3a^{2/3}} = \infty$
Therefore, at $x=0$, the tangent line is a vertical line.
$y = \sqrt[3] x~~$ has a vertical tangent line at $(0,0)$
