Chapter 4, Problem 18CR

To determine

To calculate: The average rate of change of $f\left(x\right)=x^2+3x+1$ from $1$ to 2. Use this result to find the equation of line containing $\left(1,f\left(1\right)\right)$ and $\left(2,f\left(2\right)\right)$

Answer to Problem 18CR

Solution:

The average rate of change of $f\left(x\right)=x^2+3x+1$ from $1$ to 2 is $6$ .

The equation of the secant line is $y=6x-1$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=x^2+3x+1$

Formula used:

1) The average rate of change $=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

2) Slope of a line containing points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $m=\frac{y_2-y_1}{x_2-x_1}$

3) Point slope form of a line is $y-y_1=m\left(x-x_1\right)$

Calculation:

Here, $a=1$ and $b=2$

Substitute $x=1$ in $f\left(x\right)=x^2+3x+1$

  $f\left(x=1\right)={\left(1\right)}^2+3\left(1\right)+1$

  $\Rightarrow f\left(1\right)=1+3+1$

  $\Rightarrow f\left(1\right)=5$

Now, substitute $x=2$ in $f\left(x\right)=x^2+3x+1$ ,

  $f\left(x=2\right)={\left(2\right)}^2+3\left(2\right)+1$

  $\Rightarrow f\left(2\right)=4+6+1$

  $\Rightarrow f\left(2\right)=11$

By using the formula for average rate of change

The average rate of change of a function $f\left(x\right)$ from $1$ to $2$ is $\frac{f\left(2\right)-f\left(1\right)}{2-1}$

By substituting values of $f\left(2\right)$ and $f\left(1\right)$ in $\frac{f\left(2\right)-f\left(1\right)}{2-1}$

Average rate of change $=\frac{11-5}{2-1}$

  $=\frac{6}{1}$

  $=6$

So, $\frac{f\left(2\right)-f\left(1\right)}{2-1}=6$ ----------- (1)

Thus, the average rate of change of $f\left(x\right)=x^2+3x+1$ from $1$ to 2 is $6$

Now, to find the equation of the secant line containing $\left(1,f\left(1\right)\right)$ and $\left(2,f\left(2\right)\right)$ .

First, find the slope of the secant line by using theformula of the slope of line

Slope of line $m=\frac{f\left(2\right)-f\left(1\right)}{2-1}$

From (1)

Slope of line is $m=6$

Now, substitute $m=6$ and $\left(1,f\left(1\right)\right)=\left(1,5\right)$ in the point slope form of line

  $y-5=6\left(x-1\right)$

  $\Rightarrow y-5=6x-6$

  $\Rightarrow y=6x-6+5$

  $\Rightarrow y=6x-1$

Therefore, the equation of the secant line is $y=6x-1$