Q. 59 - Calculus

Question

Suppose $f$ is a linear function with a positive slope. Show that the average rate of change of $f$on an interval $[a,b]$ is positive, and then use this fact to show that $f$ is always increasing.

Step-by-Step Solution

Answer

Ans:   the average rate of change of the function for any two instantaneous points $x1, x2$ in $[a,b]$ is given by

        $\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=\frac{y_2-y_1}{x_2-x_1}$  is positive.

Step 1. Given information.

given, 

     $f$ is a linear function with a positive slope. 

Step 2. Suppose f is a linear function with a positive slope.

The objective is to show that the average rate of change of $f$on any interval $[a,b]$ is positive, and then use this fact to show that $f$ is always increasing.

The slope of the function at any point in the interval $[a,b]$ is,

        $f^{\mathrm{\prime }}\left(x_1\right)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}$

Let  $f\left(x_1\right)=y_1\text{ and }f\left(x_2\right)=y_2$

Then, the average rate of change of the function for any two instantaneous points $x1, x2$ in $[a,b]$ is given by

         $\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=\frac{y_2-y_1}{x_2-x_1}$

Step 3. If both points are very close to each other.

Then, 

     $\frac{y_2-y_1}{x_2-x_1}=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}$

This is the average rate of change of the function.

As the slope of the function is positive,

          $\begin{array}{r}\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}>0\\ f\left(x_2\right)-f\left(x_1\right)>0\\ f\left(x_2\right)>f\left(x_1\right) \end{array}$

Thus, the function is increasing.

Hence, the average rate of change of $f$ on any interval $[a,b]$ is positive, then $f$ is always increasing.