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 126: 37


The proof is detailed below.

Work Step by Step

- The right-hand derivative of $f(x)$ at $P(0,0)$: As $x\to0^+$, $f(x)=x$. Therefore, $$\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to0^+}\frac{h-0}{h}=\lim_{h\to0^+}1=1$$ - The left-hand derivative of $f(x)$ at $P(0,0)$: As $x\to0^-$, $f(x)=x^2$. Therefore, $$\lim_{h\to0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to0^-}\frac{h^2-0}{h}=\lim_{h\to0^-}h=0$$ Since the left-hand derivative differs with the right-hand one, we conclude that $f(x)$ is not differentiable at $P$.
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