University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises - Page 127: 58


The proof is detailed in the Work step by step below.

Work Step by Step

a) $|f(x)|\le x^2$ for $-1\le x\le 1$. Consider the inequality: $$|f(x)|\le x^2$$ $$-x^2\le f(x)\le x^2$$ Since $\lim_{x\to0}(-x^2)=\lim_{x\to0}(x^2)=0$, according to the Squeeze Theorem, $\lim_{x\to0}f(x)=0$. Also, since $-x^2\le f(x)\le x^2$, for $x=0$: $$-0^2\le f(x)\le0^2$$ $$0\le f(x)\le 0$$ So, $f(x)=0$ for $x=0$. We have the derivative $f'(0)$: $$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\frac{\lim_{h\to0}f(h)-0}{h}=\frac{\lim_{h\to0}f(h)}{h}$$ Recall that $\lim_{x\to0}f(x)=0$ with $x=h$ here: $$f'(0)=\frac{0}{h}=0$$ So $f'(0)=0$. $f$ is differentiable at $x=0$. b) $f(x)=x^2\sin1/x$ for $x\ne0$ and $f(x)=0$ for $x=0$ We know that $$-1\le\sin\frac{1}{x}\le1$$ $$-x^2\le x^2\sin\frac{1}{x}\le x^2$$ $$-x^2\le f(x)\le x^2$$ $$|f(x)|\le x^2$$ As proved in a), $f$ is differentiable at $x=0$ and $f'(0)=0$.
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