Q 88. - Precalculus Enhanced with Graphing Utilities

Question

Problems 81– 88 require the following discussion of a secant line. The slope of the secant line containing the two points $(x, f(x\left)\right)$ and $(x+h,f(x+h\left)\right)$ on the graph of a function $y=f\left(x\right)$ may be given as $m_{sec}=\frac{f(x+h)-f\left(x\right)}{h}$, where $h\ne 0$.

(a) Express the slope of the secant line of each function in terms of $x$ and $h$. Be sure to simplify your answer.

(b) Find $m_{sec}$ for $h = 0.5, 0.1,$ and $0.01$ at $x=1$. What value does $m_{sec}$ approach as $h$ approaches $0$?

(c) Find the equation for the secant line at $x=1$ with $h=0.01$.

(d) Use a graphing utility to graph f and the secant line found in part (c) on the same viewing window.

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

Step-by-Step Solution

Answer

Part (a) The slope of the secant line of the given function is $m_{sec}=\frac{-h-2x}{x^2{(x+h)}^2}$.

Part (b) The value of $m_{sec}$ for $h=0.5,0.1,$ and $0.01$ are $-1.11,-1.74,$ and $-1.97$. The value of $m_{sec}$ is $-2$ when $h$ approahes $0$.

Part (c) The equation for the secant line at $x=1$ with $h=0.01$ is $y=1.97x-0.97$.

Part (d) The graph of the given function and the secant line is shown below:

Part (a) Step 1. Given information

Consider the function.

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

Part (a) Step 2. Express the slope of the secant line of the given function in terms of x and h .

The slope of secant line of the given function is as follows:

$\begin{array}{rcl}{m}_{sec}& =& \frac{f(x+h)-f\left(x\right)}{h}\\ & =& \frac{{\displaystyle \frac{1}{{(x+h)}^2}-\frac{1}{x^2}}}{h}\\ & =& \frac{\frac{x^2-(x^2+h^2+2xh)}{x^2{(x+h)}^2}}{h}\\ & =& \frac{-h^2-2xh}{h{x}^2{(x+h)}^2}\\ & =& \frac{-h-2x}{x^2{(x+h)}^2}\end{array}$

Part (b) Step 1. Determine the of m s e c for h = 0 . 5 , 0 . 1 , and 0 . 01 at . Find the value of m s e c when h approaches 0 .

From part (a), the value is $m_{sec}=\begin{array}{rcl}& & \frac{-h-2x}{x^2{(x+h)}^2}\end{array}$.

The value of $m_{sec}$ at $x=1$ with $h=0.5$ is as follows:

 $\begin{array}{rcl}{m}_{sec}& =& \frac{-0.5-2\left(1\right)}{1^2{(1+0.5)}^2}\\ & =& -\frac{2.5}{1.5^2}\\ & \approx & -1.11\end{array}$

The value of $m_{sec}$ at $x=1$ with $h=0.1$ is as follows:

 $\begin{array}{rcl}{m}_{sec}& =& \frac{-0.1-2\left(1\right)}{1^2{(1+0.5)}^2}\\ & =& -\frac{2.1}{1.1^2}\\ & \approx & -1.74\end{array}$

The value of $\begin{array}{rcl}& & m_{sec}\end{array}$ at $x=1$ with $h=0.01$ is as follows:

$\begin{array}{rcl}{m}_{sec}& =& \frac{-0.01-2\left(1\right)}{1^2{(1+0.0)}^2}\\ & =& -\frac{2.5}{1.{01}^2}\\ & \approx & -1.97\end{array}$

Thus, conclude that as $h$ approaches to $0$, $\begin{array}{rcl}& & m_{sec}\end{array}$ approaches $-2$.

Part (c) Step 1. Determine the equation for the secant line at x = 1 with h = 0 . 01 .

From part (b), $m_{sec}=-1.97$ at $x=1$ and $h=0.01$.

The equation of a straight line with point $(x_1,y_1)$ and the slope $m$ is given by $y-y_1=m(x-x_1)$.

 $\begin{array}{rcl}(1,f(1\left)\right)& =& \left(1,\frac{1}{1^2}\right)\\ & =& (1,1)\end{array}$

Thus, the equation of a line with point $(1,f(1\left)\right)$ and the slope $m=1.97$ is as follows:

$\begin{array}{rcl}y-1& =& 1.97(x-1)\\ y& =& 1.97x-1.97+1\\ y& =& 1.97x-0.97\end{array}$

Part(d) Step 1. Graph f and the secant line found in part (c) by applying a graphing utility.

The function is $f\left(x\right)=\frac{1}{x^2}$ and the secant line is $\begin{array}{rcl}y& =& 1.97x-0.97\end{array}$.

The graph of $f$ and the secant line is as follows: