Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.2 - Quadratic Functions - Exercise Set - Page 334: 108

Answer

The required expression is\[4h-2+8x\]

Work Step by Step

In order to find the value of $\frac{f\left( x+h \right)-f\left( x \right)}{h},$ calculate $f\left( x+h \right)$. $\begin{align} & f\left( x+h \right)=4{{\left( x+h \right)}^{2}}-2\left( x+h \right)+7 \\ & =4\left( {{x}^{2}}+{{h}^{2}}+2hx \right)-2x-2h+7 \\ & =4{{x}^{2}}+4{{h}^{2}}+8xh-2x-2h+7 \end{align}$ Now, putting the value $f\left( x+h \right)$ in $\frac{f\left( x+h \right)-f\left( x \right)}{h}$: $\begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{4{{x}^{2}}+4{{h}^{2}}+8xh-2x-2h+7-\left( 4{{x}^{2}}-2x+7 \right)}{h} \\ & =\frac{4{{x}^{2}}+4{{h}^{2}}+8xh-2x-2h+7-4{{x}^{2}}+2x-7}{h} \\ & =\frac{4{{h}^{2}}-2h+8xh}{h} \\ & =\frac{h\left( 4h-2+8x \right)}{h} \end{align}$ Hence, the value of $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ is $4h-2+8x$.
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