## Precalculus (6th Edition) Blitzer

The value of the expression is $5$.
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 $\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}=5$.