University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 119: 39


The graph of $y=x^{2/5}$ does not have any vertical tangents anywhere.

Work Step by Step

$$y=f(x)=x^{2/5}$$ 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: $$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{h^{2/5}-0^{2/5}}{h}=\lim_{h\to0}\frac{h^{2/5}}{h}$$ $$f'(0)=\lim_{h\to0}h^{-3/5}=\lim_{h\to0}\frac{1}{h^{3/5}}$$ As $h\to0^+$, $h^{3/5}\to0^+$, so $\lim_{h\to0^+}(1/h^{3/5})=\infty$ As $h\to0^-$, $h^{3/5}\to0^-$, so $\lim_{h\to0^-}(1/h^{3/5})=-\infty$ This means $\lim_{h\to0}(1/h^{3/5})$ does not exist. Therefore, as predicted, $f$ does not have a vertical tangent at the origin.
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