Precalculus (6th Edition) Blitzer

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

Chapter 1 - Section 1.4 - Linear Functions and Slope - Exercise Set - Page 216: 123


The value of the expression is $5$.

Work Step by Step

For $f\left( x \right)={{x}^{2}}$ , we can rewrite the expression as below: $\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}=\frac{x_{2}^{2}-x_{1}^{2}}{{{x}_{2}}-{{x}_{1}}}$ Substitute ${{x}_{1}}=1\ \text{ and }\ {{x}_{2}}=4$ to get $\begin{align} & \frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}=\frac{{{\left( 4 \right)}^{2}}-{{\left( 1 \right)}^{2}}}{4-1} \\ & =\frac{16-1}{3} \\ & =\frac{15}{3} \\ & =5 \end{align}$ Hence, the value of the expression is $\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}=5$.
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