## University Calculus: Early Transcendentals (3rd Edition)

The graph of $y=x^{5/3}-5x^{2/3}$ does not have any vertical tangents anywhere.
$$y=f(x)=x^{5/3}-5x^{2/3}$$ a) The graph is enclosed below. It does not seem to have any vertical tangents, even at the origin, because as we see from the introduction that the graph of $y=x^{2/3}$ does not have a vertical tangent at the origin. 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: $$f′(0)=\lim_{h\to0}\frac{f(h)−f(0)}{h}=\lim_{h\to0}\frac{(h^{5/3}-5h^{2/3})-(0^{5/3}-5\times0^{2/3})}{h}$$ $$f'(0)=\lim_{h\to0}\frac{h^{5/3}-5h^{2/3}}{h}=\lim_{h\to0}\Big(h^{2/3}-\frac{5}{h^{1/3}}\Big)$$ $$f'(0)=0^{2/3}-\lim_{h\to0}\frac{5}{h^{1/3}}=-\lim_{h\to0}\frac{5}{h^{1/3}}$$ - As $h\to0^+$, $h^{1/3}\to0^+$, so $\frac{5}{h^{1/3}}\to\infty$ and $\lim_{h\to0^+}\frac{5}{h^{1/3}}=\infty$ - As $h\to0^-$, $h^{1/3}\to0^-$, so $\frac{5}{h^{1/3}}\to-\infty$ and $\lim_{h\to0^-}\frac{5}{h^{1/3}}=-\infty$ This means $\lim_{h\to0}(\frac{5}{h^{1/3}})$ does not exist, and $f'(0)$ does not exist as a result. Therefore, as predicted, $f$ does not have a vertical tangent at the origin.