Answer
a. See graph.
b. vertical tangents at $x=0$ and $x=1$
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 at $x=0$: $\lim_{h\to0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0}\frac{h^{1/3}+(h-1)^{1/3}+1}{h}=\lim_{h\to0}(\frac{1}{h^{2/3}}+\frac{(h-1)^{1/3}}{h}+\frac{1}{h})=\lim_{h\to0}\frac{1}{h^{2/3}}=\infty$. thus the function has a vertical tangent at $x=0$
(ii) Evaluate the limit at $x=1$: $\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{(1+h)^{1/3}+(h)^{1/3}-1}{h}=\lim_{h\to0}(\frac{(1+h)^{1/3}}{h}+\frac{1}{h^{2/3}}-\frac{1}{h})=\lim_{h\to0}\frac{1}{h^{2/3}}=\infty$. thus the function has a vertical tangent at $x=1$
