## Precalculus (6th Edition) Blitzer

The average rate of change $f\left( x \right)={{x}^{2}}-1$ from ${{x}_{1}}=1$ to ${{x}_{2}}=2$ is 3.
Let us consider the function, $f\left( x \right)={{x}^{2}}-1$ The average rate of change of a function is defined as, “Also, assume that, $\left( {{x}_{1}},f\left( {{x}_{1}} \right) \right)$ and $\left( {{x}_{2}},f\left( {{x}_{2}} \right) \right)$ are two distinct points on the graph of a function $f\left( x \right)$. And the average rate of change of $f\left( x \right)$ from ${{x}_{1}}\text{ to }{{x}_{2}}$ is, $\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}$ ” For $f\left( x \right)$ at ${{x}_{1}}=1$ , \begin{align} & f\left( {{x}_{1}} \right)={{x}_{1}}^{2}-1 \\ & ={{1}^{2}}-1 \\ & =1-1 \\ & =0 \end{align} For $f\left( x \right)$ at ${{x}_{2}}=2$ , \begin{align} & f\left( {{x}_{2}} \right)={{x}_{2}}^{2}-1 \\ & ={{2}^{2}}-1 \\ & =4-1 \\ & =3 \end{align} Put ${{x}_{1}}=1$, ${{x}_{2}}=2$, $f\left( {{x}_{1}} \right)=0\text{ }$ and $f\left( {{x}_{2}} \right)=3$ in the formula: \begin{align} & \frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}=\frac{3-0}{2-1} \\ & =\frac{3}{1} \\ & =3 \end{align} Thus the average rate of change of $f\left( x \right)={{x}^{2}}-1$ from ${{x}_{1}}=1$ to ${{x}_{2}}=2$ is 3.