Chapter 1, Problem 8PS

  

To determine

To find: Average rate of change of f(x) from $x_1\text{ to }{x}_2$ and the secants lines through the points $(x_1{\text{,f(x}}_1\text{)) and (}{x}_2,f(x_2))$ .

Explanation of Solution

Given function is $f(x)=-x^2+4x-3$

(a) $x_1=1,x_2=2$

  $\begin{array}{l}f(x_1)=f(1)=-{(}^1+4(1)-3=0\\ f(x_2)=f(2)=-{(}^2+4(2)-3=1\end{array}$

Average rate of change= $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{1-0}{2-1}=1$

(b) $x_1=1,x_2=1.5$

  $\begin{array}{l}f(x_1)=f(1)=-{(}^1+4(1)-3=0\\ f(x_2)=f(1.5)=-{(}^{1.5}+4(1.5)-3=\frac{3}{4}\end{array}$

Average rate of change= $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{\frac{3}{4}-0}{1.5-1}=1.5$

(c) $x_1=1,x_2=1.25$

  $\begin{array}{l}f(x_1)=f(1)=-{(}^1+4(1)-3=0\\ f(x_2)=f(1.25)=-{(}^{1.25}+4(1.25)-3=\frac{7}{16}\end{array}$

Average rate of change = $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{\frac{7}{16}-0}{1.25-1}=1.75$

(d) $x_1=1,x_2=1.125$

  $\begin{array}{l}f(x_1)=f(1)=-{(}^1+4(1)-3=0\\ f(x_2)=f(1.25)=-{(}^{1.125}+4(1.125)-3=\frac{15}{64}\end{array}$

  $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{\frac{15}{64}-0}{1.125-1}=\frac{15}{8}=1.875$

(e) $x_1=1,x_2=1.0625$

  $\begin{array}{l}f(x_1)=f(1)=-{(}^1+4(1)-3=0\\ f(x_2)=f(1.0625)=-{(}^{1.0625}+4(1.0625)-3=\frac{31}{256}\end{array}$

Average rate of change = $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{\frac{31}{256}-0}{1.0625-1}=\frac{31}{16}=1.93$

(f) As the value of $x_2$ approaches to 1 average rate of change seem to approaches to 2.

(g) The equations of secant lines through the points $(x_1{\text{,f(x}}_1\text{)) and (}{x}_2,f(x_2))$ are given by

Slope of the line for $x_1=1,x_2=2$ , is $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{1-0}{2-1}=1$

Then equation of line is

  $\begin{array}{l}\frac{y-0}{x-1}=1\\ \Rightarrow y=x-1\end{array}$

Slope of the line for $x_1=1,x_2=1.5$ , is $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{\frac{3}{4}-0}{1.5-1}=1.5$

Then equation of line is

  $\begin{array}{l}\frac{y-0}{x-1}=1.5=\frac{3}{2}\\ \Rightarrow 2y=3x-3\end{array}$

Slope of the line for $x_1=1,x_2=1.25$ , is $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{\frac{7}{16}-0}{1.25-1}=1.75$

Then equation of the line is

  $\begin{array}{l}\frac{y-0}{x-1}=1.75=\frac{7}{4}\\ \Rightarrow 4y=7x-7\end{array}$

Slope of the line for, $x_1=1,x_2=1.125$ , is $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{\frac{15}{64}-0}{1.125-1}=\frac{15}{8}$

Then equation of line is

  $\begin{array}{l}\frac{y-0}{x-1}=\frac{15}{8}\\ \Rightarrow 8y=15x-15\end{array}$

Slope of the line for, $x_1=1,x_2=1.0625$ is $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{\frac{31}{256}-0}{1.0625-1}=\frac{31}{16}$

Then equation of line is

  $\begin{array}{l}\frac{y-0}{x-1}=\frac{31}{16}\\ \Rightarrow 16y=31x-31\end{array}$

(h) From (f) the slope of the given curve for the point (1, f (1)) is 2

And $f(1)=-{(1)}^2+4(1)-3=0$

Then equation is given by,

  $\begin{array}{l}\frac{y-0}{x-1}=2\\ \Rightarrow y=2x-2\end{array}$