Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises - Page 157: 47

Answer

$\dfrac{1}{(a+1)(a+h+1)}$

Work Step by Step

$f(x)=\dfrac{x}{x+1}$ Find $f(a)$ by substituting $x$ with $a$: $f(a)=\dfrac{a}{a+1}$ Find $f(a+h)$ by substituting $x$ with $a+h$: $f(a+h)=\dfrac{a+h}{a+h+1}$ We have $f(a)$ and $f(a+h)$, substitute them in the expression $\dfrac{f(a+h)-f(a)}{h}$ and simplify: $\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{a+h}{a+h+1}-\dfrac{a}{a+1}}{h}=...$ Evaluate the substraction indicated in the numerator and simplify: $...=\dfrac{\dfrac{(a+h)(a+1)-a(a+h+1)}{(a+1)(a+h+1)}}{h}=...$ $...=\dfrac{\dfrac{a^{2}+a+ah+h-a^{2}-ah-a}{(a+1)(a+h+1)}}{h}=\dfrac{\dfrac{h}{(a+1)(a+h+1)}}{h}=...$ $...=\dfrac{1}{(a+1)(a+h+1)}$
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