Chapter 3.9, Problem 32WE

To determine

To prove:The value of $\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=m$ for the given information.

Explanation of Solution

Given information:The value of $f\left(x\right)=mx+b$ and $x_1\ne x_2$ .

Formula used:

The equation of a linear function is given by

  $f\left(x\right)=mx+b,\text{where}m\text{is}\text{the}\text{slope}\text{of}\text{the}\text{function}$ .

Proof:

The linear function is $f\left(x\right)=mx+b$   ........(1)

Put $x=x_1\text{in}\text{equation}\left(1\right)$   $f\left(x_1\right)=m{x}_1+b\mathrm{........}\left(2\right)$

Put   $x=x_2\text{in}\text{equation}\left(1\right)$   $f\left(x_2\right)=m{x}_2+b\mathrm{........}\left(3\right)$

Then using the results of equation (2) & (3)

  $\begin{array}{c}\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=\frac{\left(m{x}_2+b\right)-\left(m{x}_1+b\right)}{\left(x_2-x_1\right)}\\ =\frac{m{x}_2+b-m{x}_1-b}{\left(x_2-x_1\right)}\\ =\frac{m\left(x_2-x_1\right)}{\left(x_2-x_1\right)}=m\\ \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=m\end{array}$

Hence proved that thevalue of $\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=m$ .