Answer
The average rate of change $f\left( x \right)={{x}^{2}}-1$ from ${{x}_{1}}=1$ to ${{x}_{2}}=2$ is 3.
Work Step by Step
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.
