## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 1 - Section 1.5 - More on Slope - Exercise Set - Page 226: 17

#### Answer

The average rate of change of $f\left( x \right)=\sqrt{x}$ from ${{x}_{1}}=4\text{ to }{{x}_{2}}=9$ is $\frac{1}{5}$.

#### Work Step by Step

Consider the provided function $f\left( x \right)=\sqrt{x}$. The value of function $f\left( x \right)=\sqrt{x}$ at ${{x}_{1}}=4$ is \begin{align} & f\left( 4 \right)=\sqrt{\left( 4 \right)} \\ & =2 \end{align} The value of function $f\left( x \right)=\sqrt{x}$ at ${{x}_{2}}=9$ is \begin{align} & f\left( 9 \right)=\sqrt{\left( 9 \right)} \\ & =3 \end{align} The average rate of change of $f$ from ${{x}_{1}}=4\text{ to }{{x}_{2}}=9$ is \begin{align} & \frac{\Delta y}{\Delta x}=\frac{f\left( 9 \right)-f\left( 4 \right)}{9-4} \\ & =\frac{3-2}{9-4} \\ & =\frac{1}{5} \end{align} Thus, the average rate of change of $f\left( x \right)=\sqrt{x}$ from ${{x}_{1}}=4\text{ to }{{x}_{2}}=9$ is $\frac{1}{5}$.

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