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: 44

Answer

The graph of $y=x^{5/3}-5x^{2/3}$ does not have any vertical tangents anywhere.
1537614357

Work Step by Step

$$y=f(x)=x^{5/3}-5x^{2/3}$$ 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.
Small 1537614357
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.