Answer
a) $g'(0)$ does not exist.
b) The slope of the tangent is undefined.
Work Step by Step
$a) g'(0)$ does not exist.
The graph of $g(x)$ is attached. At g(0), the slope of the graph becomes vertical, and since the slope of a vertical line is undefined, the differentiated function $g'(x)$ does not take up any value at $x = 0$.
Arithmetic Proof:
$\sqrt[3] x = x^{\frac{1}{3}}$
Using Power rule:
$\implies \frac{d(x^{\frac{1}{3}})}{dx}= \frac{1}{3}\times x^{\frac{1}{3}-1}\\=\frac{x^{-\frac{2}{3}}}{3} = \frac{1}{\sqrt[3] {(x^2)}\times3}$
If $x = 0$, the denominator would be 0, therefore $g'(0)$ does not exist.
$b)$ The derivative of a function at any given point is defined as the slope of the line that's tangential to that point on the function. Here, since the tangent line is vertical, the slope of it does not exist, and hence, the derivative function does not exist at $x = 0$.
