Q 81. - 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.

$f\left(x\right)=2x+5$

Step-by-Step Solution

Answer

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

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

Part (c) The equation for the secant line at $x=1$ with $h=0.01$ is $y=2x+5$.

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

Part (a) Step 1. Given information

Consider the function.

$f\left(x\right)=2x+5$

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:

$\begin{array}{rcl}{m}_{sec}& =& \frac{f(x+h)-f\left(x\right)}{h}\\ & =& \frac{2(x+h)+5-(2x+5)}{h}\\ & =& \frac{2x+2h+5-2x-5}{h}\\ & =& \frac{2h}{h}\\ & =& 2\end{array}$

Part (b) Step 1. Determine the value of m s e c for h   =   0 . 5 ,   0 . 1 , and 0 . 01 at x   =   1 . Then, find the value of src="data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHhtbG5zOndycz0iaHR0cDovL3d3dy53aXJpcy5jb20veG1sL21hdGhtbC1leHRlbnNpb24iIGhlaWdodD0iMjQiIHdpZHRoPSIzNSIgd3JzOmJhc2VsaW5lPSIxNSI%2BPCEtLU1hdGhNTDogPG1hdGggeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzE5OTgvTWF0aC9NYXRoTUwiPjxtdGFibGUgY29sdW1uYWxpZ249InJpZ2h0IGNlbnRlciBsZWZ0IiBjb2x1bW5zcGFjaW5nPSIwcHgiPjxtdHI%2BPG10ZC8%2BPG10ZC8%2BPG10ZD48bXN1Yj48bWk%2BbTwvbWk%2BPG1yb3c%2BPG1pPnM8L21pPjxtaT5lPC9taT48bWk%2BYzwvbWk%2BPC9tcm93PjwvbXN1Yj48L210ZD48L210cj48L210YWJsZT48L21hdGg%2BLS0%2BPGRlZnM%2BPHN0eWxlIHR5cGU9InRleHQvY3NzIi8%2BPC9kZWZzPjx0ZXh0IGZvbnQtZmFtaWx5PSJBcmlhbCIgZm9udC1zaXplPSIxNiIgZm9udC1zdHlsZT0iaXRhbGljIiB0ZXh0LWFuY2hvcj0ibWlkZGxlIiB4PSI2LjUiIHk9IjE1Ij5tPC90ZXh0Pjx0ZXh0IGZvbnQtZmFtaWx5PSJBcmlhbCIgZm9udC1zaXplPSIxMiIgZm9udC1zdHlsZT0iaXRhbGljIiB0ZXh0LWFuY2hvcj0ibWlkZGxlIiB4PSIxNy41IiB5PSIyMCI%2BczwvdGV4dD48dGV4dCBmb250LWZhbWlseT0iQXJpYWwiIGZvbnQtc2l6ZT0iMTIiIGZvbnQtc3R5bGU9Iml0YWxpYyIgdGV4dC1hbmNob3I9Im1pZGRsZSIgeD0iMjMuNSIgeT0iMjAiPmU8L3RleHQ%2BPHRleHQgZm9udC1mYW1pbHk9IkFyaWFsIiBmb250LXNpemU9IjEyIiBmb250LXN0eWxlPSJpdGFsaWMiIHRleHQtYW5jaG9yPSJtaWRkbGUiIHg9IjMwLjUiIHk9IjIwIj5jPC90ZXh0Pjwvc3ZnPg%3D%3D" localid="1647174099392" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/6224c363-9582-455d-9b2a-9577eb242f1f.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220315%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220315T122409Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=58a7de762c6d618de3945ddbdb689f816f58762872da5d678e80f30775aba6b2" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/6224c363-9582-455d-9b2a-9577eb242f1f.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220315%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220315T115758Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=d712dba90515fffa94de92dbaf257708ed8dd96f507ab0d0e63f9a6106e99093" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/6224c363-9582-455d-9b2a-9577eb242f1f.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220315%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220315T115339Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=9dc599d08d48d48844c0f38a6c5b33788ebff5372f16da4f8e5e61a0c05c4906" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/6224c363-9582-455d-9b2a-9577eb242f1f.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220315%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220315T113820Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=1d57150120a9c9008af8fe994082c6cf1984b4aa4162ed16338ae03d4f583bb4" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/6224c363-9582-455d-9b2a-9577eb242f1f.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220314%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220314T125425Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=44ca68b806b2f93299db239445eb1ec829f434f4c0f2b87d011849e30bc757f3" m s e c when h approaches 0 .

From part (a), the value is $\begin{array}{rcl}{m}_{sec}& =& 2\end{array}$ for all values of $h$ since $\begin{array}{rcl}& & m_{sec}\end{array}$ is constant in $h$.

Therefore, the value of $m_{sec}$ for $h=0.5,0.1,$ and $0.01$ at $x=1$ are $2,2,$ and $2$.

Thus, conclude that as $h$ approaches to $0$, $m_{sec}$ approaches $2$.

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

From part (b), $m_{sec}=2$ 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 $y-y_1=m(x-x_1)$.

$\begin{array}{rcl}(1,f(1\left)\right)& =& (1,2(1)+5)\\ & =& (1,2+5)\\ & =& (1,7)\end{array}$

Thus, the equation of a line with point $(1,f(1\left)\right)$ and the slope $m=2$ is as follows:

$\begin{array}{rcl}y-7& =& 2(x-1)\\ y-7& =& 2x-2\\ y& =& 2x-2+7\\ y& =& 2x+5\end{array}$

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

The function is $f\left(x\right)=2x+5$ and the secant line is $y=2x+5$.

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