Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.8 - The Derivative as a Function - 2.8 Exercises - Page 164: 57

Answer

(a) $f'(a) = \frac{1}{3a^{2/3}}$ (b) $f'(0)$ does not exist. (c) $y = \sqrt[3] x~~$ has a vertical tangent line at $(0,0)$

Work Step by Step

(a) $f(x) = \sqrt[3] x$ We can find $f'(a)$: $\lim\limits_{x \to a}\frac{f(x)-f(a)}{x-a}$ $= \lim\limits_{x \to a}\frac{\sqrt[3] x-\sqrt[3] a}{x-a}$ $= \lim\limits_{x \to a}\frac{\sqrt[3] x-\sqrt[3] a}{(x^{1/3})^3-(a^{1/3})^3}$ $= \lim\limits_{x \to a}\frac{\sqrt[3] x-\sqrt[3] a}{(x^{1/3}-a^{1/3})(x^{2/3}+x^{1/3}a^{1/3}+a^{2/3})}$ $= \lim\limits_{x \to a}\frac{1}{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}}$ $= \frac{1}{a^{2/3}+a^{1/3}a^{1/3}+a^{2/3}}$ $= \frac{1}{3a^{2/3}}$ (b) We can try to find $f'(0)$: $f'(a) = \frac{1}{3a^{2/3}}$ $f'(0) = \frac{1}{3(0)^{2/3}}$ This expression does not exist when $a = 0$ Therefore, $f'(0)$ does not exist. (c) Note that $f(0) = 0$ The slope of a tangent line at a certain point is equal to the slope of the graph at that point. The slope of a graph at point $a$ is $f'(a)$ We can find $f'(a)$ as $a$ approaches 0: $\lim\limits_{a \to 0}f'(a) = \lim\limits_{a \to 0}\frac{1}{3a^{2/3}} = \infty$ Therefore, at $x=0$, the tangent line is a vertical line. $y = \sqrt[3] x~~$ has a vertical tangent line at $(0,0)$
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