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?

Consider the function.

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

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 role="math" localid="1647153942887" $a=1$and role="math" localid="1647154803807" $b=2$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 $3$.

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 $2.5$.

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 $2.1$.

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 $2.01$.

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 $2.001$.

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

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

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 $2$.