Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 40

Answer

$f(x)=x^6$ and $a=2$

Work Step by Step

According to definition, $$f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$$ Therefore, at a number a, the derivative would be $$f'(a)=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Now we look at the given formula: $$f'(a)=\lim\limits_{h\to0}\frac{(2+h)^6-64}{h}$$ $$f'(a)=\lim\limits_{h\to0}\frac{(2+h)^6-2^6}{h}$$ Here, from the definition, we can deduce that $a=2$. That means $f(a)=2^6$. Therefore, $f(x)=x^6$
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