Precalculus (6th Edition) Blitzer

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

Chapter 7 - Cumulative Review Exercises - Page 880: 30


The required solution is $\frac{f\left( x+h \right)-f\left( x \right)}{h}=2h+4x-1$.

Work Step by Step

We have the functions $f\left( x \right)=2{{x}^{2}}-x-1$ and $ g\left( x \right)=1-x $: Therefore, compute $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ as given below: $\begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{2{{\left( x+h \right)}^{2}}-\left( x+h \right)-1-\left( 2{{x}^{2}}-x-1 \right)}{h} \\ & =\frac{2{{\left( x+h \right)}^{2}}-\left( x+h \right)-1-\left( 2{{x}^{2}}-x-1 \right)}{h} \\ & =\frac{2{{h}^{2}}+4xh-h}{h} \\ & =2h+4x-1 \end{align}$ Thus, $\frac{f\left( x+h \right)-f\left( x \right)}{h}=2h+4x-1$.
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