Q 80. - Precalculus Enhanced with Graphing Utilities

Question

For the function $f\left(x\right)=x^2$, compute each average rate of change:

(a) From $1$ to $2$

(b) From $1$ to $1.5$

(c) From $1$ to $1.1$

(d) From $1$ to $1.01$

(e) From $1$ to $1.001$

(f) Use a graphing utility to graph each of the secant lines along with $f$.

(g) What do you think is happening to the secant lines?

(h) What is happening to the slopes of the secant lines? Is there some number that they are getting closer to? What is that number?

Step-by-Step Solution

Answer

Part (a). The average rate of change from $1$ to $2$ is $3$.

Part (b). The average rate of change from $1$ to $1.5$ is $2.5$.

Part (c). The average rate of change from $1$ to $1.1$ is $2.1$.

Part (d). The average rate of change from $1$ to $1.01$ is $2.01$.

Part (e). The average rate of change from $1$ to $1.001$ is $2.001$.

Part (f). The graph of the secant lines is shown below:

Part (g). The secant lines is decreasing and approaching closer to the straight line $y=2x$, that is, at the point $(1,1)$.

Part (h). The slopes of the secant lines is decreasing and getting closer to $(1,1)$. The slope is approaching the value of $2$.

Part (a) Step 1. Given information

Consider the function.

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

Part (a) Step 2. Determine the average rate of change from 1 to 2 .

The average rate of change from $a$ to $b$ is defined as folllows:

$\frac{∆y}{∆x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}\dots \left(1\right)$, where $a\ne b$.

Substitute $a=1$ and $b=2$ into $\left(1\right)$.

$\frac{∆y}{∆x}=\frac{f\left(2\right)-f\left(1\right)}{2-1}$

Since $f\left(x\right)=x^2$, the average rate of change is as follows:

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{2^2-1^2}{1}\\ & =& \frac{4-1}{1}\\ & =& 3\end{array}$

Thus, the average rate of change is $3$.

Part (b) Step 1. Determine the average rate of change from 1 to 1 . 5 .

Substitute $a=1$ and $b=1.5$ into $\left(1\right)$.

$\frac{∆y}{∆x}=\frac{f(1.5)-f\left(1\right)}{1.5-1}$

Since $f\left(x\right)=x^2$, the average rate of change is as follows:

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{1.5^2-1^2}{1.5-1}\\ & =& \frac{2.25-1}{0.5}\\ & =& \frac{1.25}{0.5}\\ & =& 2.5\end{array}$

Thus, the average rate of change is $2.5$.

Part (c) Step 1. Determine the average rate of change from 1 to 1 . 1 .

Substitute $a=1$ and $b=1.1$ into $\left(1\right)$.

$\frac{∆y}{∆x}=\frac{f(1.1)-f\left(1\right)}{1.1-1}$

Since $f\left(x\right)=x^2$, the average rate of change is as follows:

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{1.1^2-1^2}{1.1-1}\\ & =& \frac{1.21-1}{0.1}\\ & =& \frac{0.21}{0.1}\\ & =& 2.1\end{array}$

Thus, the average rate of change is $2.1$.

Part (d) Step 1. Determine the average rate of change from 1 to 1 . 01 .

Substitute $a=1$ and $b=1.01$ into $\left(1\right)$.

$\frac{∆y}{∆x}=\frac{f(1.01)-f\left(1\right)}{1.01-1}$

Since $f\left(x\right)=x^2$, the average rate of change is as follows:

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{1.{01}^2-1^2}{1.01-1}\\ & =& \frac{1.0201-1}{0.01}\\ & =& \frac{0.0201}{0.01}\\ & =& 2.01\end{array}$

Thus, the average rate of change is $2.01$.

Part (e) Step 1. Determine the average rate of change from 1 to src="data:image/svg+xml;base64,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" role="math" localid="1647154673135" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/49417c61-f49a-4b73-8dbe-66e015c12762.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220315%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220315T115425Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=e519b333e9ecdd4d06fa7dc1c1652aabfbfb295c990cd8c612b8c42ad1460de0" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/49417c61-f49a-4b73-8dbe-66e015c12762.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220313%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220313T114926Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=24593916bca098b27cd1dcbab31a1741a0f916ae9c9436e4a4686590f1b256d3" src="https://studysmarter-mediafiles.s3.amazonaws.com/media/textbook-exercise-images/49417c61-f49a-4b73-8dbe-66e015c12762.svg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIA4OLDUDE42UZHAIET%2F20220313%2Feu-central-1%2Fs3%2Faws4_request&X-Amz-Date=20220313T111006Z&X-Amz-Expires=90000&X-Amz-SignedHeaders=host&X-Amz-Signature=4115f2a3d6149c946b538abd790256eb90b5239fc27436ba13122043e666c9c6" 1 . 001 .

Substitute $a=1$ and $b=1.001$ into $\left(1\right)$.

$\frac{∆y}{∆x}=\frac{f(1.001)-f\left(1\right)}{1.001-1}$

Since $f\left(x\right)=x^2$, the average rate of change is as follows:

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{1.{001}^2-1^2}{1.001-1}\\ & =& \frac{1.002001-1}{0.001}\\ & =& \frac{0.0201}{0.01}\\ & =& 2.001\end{array}$

Thus, the average rate of change is $2.001$.

Part (f) Step 1. Apply the graphing utility to graph each of the secant lines along with f .

The graph of each of the secant lines along with the given function $f$ is shown below:

Part (g) Step 1. Determine the status of the secant lines.

From the graph, observe that the secant lines is decreasing and approaching closer to the straight line $y=2x$, that is, at the point $(1,1)$.

Part (h) Step 1. Determine the status of the slopes of the secant lines and find if there is any number they are getting closer to?. Determine the number.

From the graph, observe that the slopes of the secant lines is decreasing and approaching closer to the point $(1,1)$.

This implies that the slope is approaching the value of $2$.