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: 82


The difference quotient for the provided function is \[-2x-h-3\].

Work Step by Step

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