Precalculus (6th Edition) Blitzer

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

Chapter 1 - Section 1.3 - More on Functions and Their Graphs - Exercise Set - Page 198: 80


The difference quotient for the provided function is \[6x+3h+1\].

Work Step by Step

Consider the provided function: $f\left( x \right)=3{{x}^{2}}+x+5$. Now, substitute $x=x+h$ in the above equation to find $f\left( x+h \right)$ That is, $\begin{align} & f\left( x+h \right)=3{{\left( x+h \right)}^{2}}+\left( x+h \right)+5 \\ & =3{{x}^{2}}+6xh+3{{h}^{2}}+x+h+5 \end{align}$ Now, apply the difference quotient formula, $\begin{align} & \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{3{{x}^{2}}+6xh+3{{h}^{2}}+x+h+5-\left( 3{{x}^{2}}+x+5 \right)}{h} \\ & =\frac{3{{x}^{2}}+6xh+3{{h}^{2}}+x+h+5-3{{x}^{2}}-x-5}{h} \\ & =\frac{6xh+3{{h}^{2}}+h}{h} \\ & =\frac{h\left( 6x+3h+1 \right)}{h} \end{align}$ Further solve and get, $\frac{f\left( x+h \right)-f\left( x \right)}{h}=6x+3h+1$ Hence, the difference quotient for the provided function is $6x+3h+1$.
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