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: 59

Answer

This expression represents the slope of the tangent of the function $f(x)=x^4$ at $a=2$ and its value is $m_{tan}=32.$

Work Step by Step

The definition of the slope of the tangent at $a$ is given by $$m_{tan}=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}.$$ We see that if we take $f(x)=x^4$ and $a=2$ (and also we calculate $f(a)=f(2)=2^4=16$) the upper expression becomes the expression for the slope of $f(x)=x^4$ at $a=2$. Let us calculate this slope. We will use that $c^4-d^4=(c^2+d^2)(c^2-d^2)=(c^2+d^2)(c+d)(c-d):$ $$m_{tan}=\lim_{h\to0}\frac{(2+h)^4-16}{h}=\lim_{h\to0}\frac{(2+h)^4-2^4}{h}=\lim_{h\to0}\frac{((2+h)^2+2^2)(2+h+2)(2+h-2)}{h}=\lim_{h\to0}\frac{((2+h)^2+4)(4+h)h}{h}=\lim_{h\to0}((2+h)^2+4)(4+h)=((2+0)^2+4)(4+0)=32.$$
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