Answer
The graph of the given function has a vertical tangent at $x=1$.
Work Step by Step
$$y=f(x)=x^{2/3}-(x-1)^{1/3}$$
a) The graph is enclosed below. From the introduction, we can guess that the graph does not seem to have a vertical tangent at $x=0$, but it appears to have a vertical tangent at $x=1$. We will test this in part b).
b) The graph of $f(x)$ has a vertical tangent at $x$ when
1) $f$ is continuous at $x$
2) $$f′(x)=\lim_{h\to0}\frac{f(x+h)−f(x)}{h}=\pm\infty$$
Here, we can see that:
$$f′(1)=\lim_{h\to0}\frac{f(h+1)−f(1)}{h}$$
$$f'(1)=\lim_{h\to0}\frac{\Big((h+1)^{2/3}-(h+1-1)^{1/3}\Big)-\Big(1^{2/3}-(1-1)^{1/3}\Big)}{h}$$
$$f'(1)=\lim_{h\to0}\frac{\Big((h+1)^{2/3}-h^{1/3}\Big)-(1-0)}{h}$$
$$f'(1)=\lim_{h\to0}\frac{(h+1)^{2/3}-h^{1/3}-1}{h}=\lim_{h\to0}\frac{(h+1)^{2/3}-1}{h}-\lim_{h\to0}\frac{h^{1/3}}{h}$$
$$f'(1)=A-B$$
$$A=\lim_{h\to0}\frac{(h+1)^{2/3}-1}{h}$$
Multiply both numerator and denominator by $(h+1)^{4/3}+(h+1)^{2/3}+1$:
$$A=\lim_{h\to0}\frac{\Big((h+1)^{2/3}-1\Big)\Big((h+1)^{4/3}+(h+1)^{2/3}+1\Big)}{h\Big((h+1)^{4/3}+(h+1)^{2/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{(h+1)^2-1}{h\Big((h+1)^{4/3}+(h+1)^{2/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{h^2+2h+1-1}{h\Big((h+1)^{4/3}+(h+1)^{2/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{h^2+2h}{h\Big((h+1)^{4/3}+(h+1)^{2/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{h+2}{(h+1)^{4/3}+(h+1)^{2/3}+1}$$ $$A=\frac{0+2}{(0+1)^{4/3}+(0+1)^{2/3}+1}=\frac{2}{3}$$
$$B=\lim_{h\to0}\frac{h^{1/3}}{h}=\lim_{h\to0}\frac{1}{h^{2/3}}$$
- As $h\to0^+$ or $h\to0^-$, $h^{2/3}$ always approaches $0^+$, so $\lim_{h\to0}\frac{1}{h^{2/3}}=\infty$.
In conclusion, $$f'(1)=A-B=\frac{2}{3}-\infty=-\infty$$
This means $f$ has a vertical tangent at $x=1$ as predicted.
