Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 2 - Problems Plus - Problems - Page 172: 14

Answer

$f(0) = 0$ $f'(0) = 0$

Work Step by Step

$\vert f(x) \vert \leq x^2~~$ for all $x$ We can find $f(0)$: $\vert f(0) \vert \leq 0^2 = 0$ Therefore, $f(0) = 0$ We can find $f'(0)$: $\vert f'(0) \vert = \vert \lim\limits_{x \to 0} \frac{f(x)-f(0)}{x-0}\vert$ $\vert f'(0) \vert = \vert \lim\limits_{x \to 0} \frac{f(x)}{x}\vert \leq \vert \lim\limits_{x \to 0} \frac{x^2}{x}\vert = \vert \lim\limits_{x \to 0} ~x \vert = 0$ Therefore, $f'(0) = 0$
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