Q 83. - 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 width="10" style="max-width: none; vertical-align: -5px;" $f$ and the secant line found in part (c) on the same viewing window.

$f\left(x\right)=x^2+2x$

Step-by-Step Solution

Answer

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

Part (b) The value of $m_{sec}$ for $h=0.5,0.1,$ and $0.01$ are $4.5,4.1,$ and $4$. The value of $m_{sec}$ is $4$ when $h$ approahes $0$.

Part (c) The equation for the secant line at $x=1$ with $h=0.01$ is $y=4.01x-1.01$.

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

Part (a) Step 1. Given information

Consider the function.

$f\left(x\right)=x^2+2x$

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:

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

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

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

The value of $\begin{array}{rcl}& & m_{sec}\end{array}$ at $x=1$ with $h=0.5$ is as follows:

$\begin{array}{rcl}{m}_{sec}& =& 0.5+2\left(1\right)+2\\ & =& 4.5\end{array}$

The value of $\begin{array}{rcl}& & m_{sec}\end{array}$ at $x=1$ with $h=0.1$ is as follows:

$\begin{array}{rcl}{m}_{sec}& =& 0.1+2\left(1\right)+2\\ & =& 4.1\end{array}$

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

$\begin{array}{rcl}{m}_{sec}& =& 0.01+2\left(1\right)+2\\ & =& 4.01\end{array}$

Thus, conclude that as $h$ approaches to $0$, $\begin{array}{rcl}& & m_{sec}\end{array}$ approaches $4$.

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

From part (b), $\begin{array}{rcl}{m}_{sec}& =& 4.01\end{array}$ 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 $y-y_1=m(x-x_1)$.

 $\begin{array}{rcl}(1,f(1\left)\right)& =& (1,{\left(1\right)}^2+2(1\left)\right)\\ & =& (1,1+2)\\ & =& (1,3)\end{array}$

Thus, the equation of a line with point $(1,f(1\left)\right)$ and the slope $m=4.01$ is as follows:

$\begin{array}{rcl}y-3& =& 4.01(x-1)\\ y-3& =& 4.01x-4.01\\ y& =& 4.01x-4.01+3\\ y& =& 4.01x-1.01\end{array}$

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

The function is $f\left(x\right)=x^2+2x$ and the secant line is $y=4.01x-1.01$.

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