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: 42

Answer

The graph of the given function has a vertical tangent at the origin.
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Work Step by Step

$$y=f(x)=x^{3/5}$$ a) The graph is enclosed below. From the introduction, we can guess that the graph appears to have vertical tangent at the origin. 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) $$f′(x)=\lim_{h\to0}\frac{f(x+h)−f(x)}{h}=\pm\infty$$ Considering $f'(0)$, we have: $$f′(0)=\lim_{h\to0}\frac{f(h)−f(0)}{h}=\lim_{h\to0}\frac{h^{3/5}-0^{3/5}}{h}=\lim_{h\to0}\frac{h^{3/5}}{h}$$ $$f'(0)=\lim_{h\to0}\frac{1}{h^{2/5}}$$ As $h\to0^+$ or $h\to0^-$, $h^{2/5}$ always approaches $0^+$. Therefore, $$f'(0)=\lim_{h\to0}\frac{1}{h^{2/5}}=\infty$$ This means $f(x)$ has a vertical tangent at the origin as predicted.
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