Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.1 Introducing the Derivative - 3.1 Execises - Page 134: 63


For the derivative to be continuous at $x=1$ we have to choose $a=4$.

Work Step by Step

By definition of continuity of the derivative at $x=a$, $f'$ is continuous if $$\lim_{x\to a^+}f'(x)=\lim_{x\to a^-}f'(x)=f'(a).$$ We will first find the derivative on the both sides of $x=1$. If $x\leq1$ then $$f'(x)=\lim_{h\to0}\frac{2(x+h)^2-2x^2}{h}=\lim_{h\to0}\frac{2(x^2+h^2+2xh)-2x^2}{h}=\lim_{h\to0}\frac{2x^2+2h^2+4xh-2x^2}{h}=\lim_{h\to0}\frac{2h^2+4xh}{h}=\lim_{h\to0}(2h+4x)=2\cdot0+4x=4x.$$ If $x>1$ then $$f'(x)=\lim_{h\to0}\frac{a(x+h)-2-(ax-2)}{h}=\lim_{h\to0}\frac{ax+ah-2-ax+2}{h}=\lim_{h\to0}\frac{ah}{h}=\lim_{h\to0}a=a.$$ Now we will find the one-sided limits. Because when $x\leq1$ $f'(x)=4x$ we have $$\lim_{x\to 1^{+}}4x=4\cdot1=4.$$ When $x>1$ then $f'(x)=a$ so we have $$\lim_{x\to1^-}a=a$$. For those one-sided limits to be equal we see that $a=4$. Then also $f'(1)=4$ so when $a=4$ the derivative is continuous.
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