Answer
If $\left( {{x}_{1}},f\left( {{x}_{1}} \right) \right)\text{ and }\left( {{x}_{2}},f\left( {{x}_{2}} \right) \right)$ are distinct points on the graph of a function $f$ , the average rate of change from ${{x}_{1}}$ to ${{x}_{2}}$ is $\frac{\Delta y}{\Delta x}=$ $\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}\text{ }$.”
Work Step by Step
The average rate of change of $f$ is given as
$\frac{\Delta y}{\Delta x}=\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$
The mathematical symbol $\frac{\Delta y}{\Delta x}$ represents the change in y divided by the change in $x$.
For example, consider the function $y={{x}^{3}}$ at ${{x}_{1}}=0\text{ to }{{x}_{2}}=2$.
The rate of change is calculated by the substitution of the value as ${{x}_{1}}=0\text{ to }{{x}_{2}}=2$ and $f\left( {{x}_{1}} \right)=0\text{ and }f\left( {{x}_{2}} \right)=8$.
$\begin{align}
& \frac{\Delta y}{\Delta x}=\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)} \\
& =\frac{8-0}{2-0} \\
& =4
\end{align}$
Thus, the average rate of change of $f\left( x \right)={{x}^{3}}$ from ${{x}_{1}}=0\text{ to }{{x}_{2}}=2$ is $4$.
