Precalculus (6th Edition) Blitzer

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

Chapter 7 - Section 7.6 - Linear Programming - Exercise Set - Page 874: 39


The value is, $10x+5h-6$.

Work Step by Step

The function is given below, $f\left( x \right)=5{{x}^{2}}-6x+1$ (I) Substitute $x=x+h$ into equation (I): $\begin{align} & f\left( x+h \right)=5{{\left( x+h \right)}^{2}}-6\left( x+h \right)+1 \\ & =5\left( {{x}^{2}}+2xh+{{h}^{2}} \right)-6x-6h+1 \\ & =5{{x}^{2}}+10xh+5{{h}^{2}}-6x-6h+1 \end{align}$ Then, calculate $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ (II) Substitute the value of $f\left( x+h \right)$ into equation (II). So, $\begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{5{{x}^{2}}+10xh+5{{h}^{2}}-6x-6h+1-5{{x}^{2}}+6x-1}{h} \\ & =\frac{10xh+5{{h}^{2}}-6h}{h} \\ & =\frac{h\left( 10x+5h-6 \right)}{h} \end{align}$ Finally, $\frac{f\left( x+h \right)-f\left( x \right)}{h}=10x+5h-6$ , $h\ne 0$ Hence, $\frac{f\left( x+h \right)-f\left( x \right)}{h}=10x+5h-6$ where $h\ne 0$.
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