Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 1 - Graphs - 2.1 Functions - 2.1 Assess Your Understanding - Page 58: 90

Answer

$\dfrac{1}{\sqrt {x+h+1}+\sqrt {x+1}}$

Work Step by Step

Step $1$. Given $f(x)=\sqrt {x+1}$, we have $f(x+h)=\sqrt {x+h+1}$ Step $2$. $f(x+h)-f(x)=\sqrt {x+h+1}-\sqrt {x+1}$ Step $3$. (Rationalize the numerator as suggested.) $\begin{align*} \dfrac{f(x+h)-f(x)}{h}&=\frac{\sqrt {x+h+1}-\sqrt {x+1}}{h}\\ &=\frac{\sqrt {x+h+1}-\sqrt {x+1}}{h}\cdot\frac{\sqrt {x+h+1}+\sqrt {x+1}}{\sqrt {x+h+1}+\sqrt {x+1}}\\ &=\frac{x+h+1-x-1}{h(\sqrt {x+h+1}+\sqrt {x+1})}\\ &=\frac{h}{h(\sqrt {x+h+1}+\sqrt {x+1})}\\ &=\frac{1}{\sqrt {x+h+1}+\sqrt {x+1}} \end{align*}$
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