Answer
a. See graph.
b. vertical tangent at $x=1$
Work Step by Step
a. See graph; it appears to have a vertical tangent at $x=0$ and $x=1$
b. (i) Evaluate the limit for $x=0$: $\lim_{h\to0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0}\frac{h^{2/3}-(h-1)^{1/3}+(-1)^{1/3}}{h}=\lim_{h\to0}(\frac{1}{h^{1/3}}- \frac{(h-1)^{1/3}}{h}-\frac{1}{h} )=\lim_{h\to0}(\frac{1}{h^{1/3}} )$. As this limit has opposite signs when $h\to0^+$ vs $h\to0^-$, the function does not have a vertical tangent at $x=0$.
(ii) Evaluate the limit for $x=1$: $\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{(1+h)^{2/3}-(h)^{1/3}-(1)^{2/3}}{h}=\lim_{h\to0}(\frac{(1+h)^{2/3}}{h}-\frac{1}{h^{2/3}}-\frac{1}{h} )=\lim_{h\to0}(-\frac{1}{h^{2/3}})$.
As this limit is the same for $h\to0^+$ and $h\to0^-$, the function has a vertical tangent at $x=1$
