Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1163: 66


The value of $\frac{f\left( 2+h \right)-f\left( 2 \right)}{h}$ is $ h+5$.

Work Step by Step

Consider the function $ f\left( x \right)={{x}^{2}}+x $, Find the value of $\frac{f\left( 2+h \right)-f\left( 2 \right)}{h}$ using the definition of the function. $\begin{align} & \frac{f\left( 2+h \right)-f\left( 2 \right)}{h}=\frac{{{\left( 2+h \right)}^{2}}+\left( 2+h \right)-\left( {{2}^{2}}+2 \right)}{h} \\ & =\frac{4+{{h}^{2}}+4h+2+h-\left( 4+2 \right)}{h} \\ & =\frac{{{h}^{2}}+5h+6-6}{h} \\ & =\frac{{{h}^{2}}+5h}{h} \end{align}$ Factoring out h from the numerator, $\begin{align} & \frac{f\left( 2+h \right)-f\left( 2 \right)}{h}=\frac{h\left( h+5 \right)}{h} \\ & =h+5 \end{align}$ Thus, the value of $\frac{f\left( 2+h \right)-f\left( 2 \right)}{h}$ is $ h+5$.
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