Show that a constant function $f\left(x\right)=b$has an average rate of change of $0$. Evaluate the average rate of change of $y=\sqrt{4-x^2}$on the interval $[-2,2]$. Explain how this can happen.

The functions are $f\left(x\right)=b$ and $f\left(x\right)=\sqrt{4-x^2}$.

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 $x_1$for $a$and $x_2$for $b$into $\left(1\right)$.

$\frac{∆y}{∆x}=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\dots \left(2\right)$

Since $f\left(x\right)=b$, the values are as follows:

$\begin{array}{rcl}f\left(x_2\right)& =& b\\ f\left(x_1\right)& =& b\end{array}$

From $\left(2\right)$, the average rate of change is as follows:

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{b-b}{x_2-x_1}\\ & =& \frac{0}{x_2-x_1}\\ & =& 0\end{array}$

Thus, it is proved that a constant function $f\left(x\right)=b$ has an average rate of change of $0$.

Assume $y=f\left(x\right)$.

So, the function is $f\left(x\right)=\sqrt{4-x^2}\dots \left(3\right)$.

To find the value of localid="1647405846259" $f\left(x\right)$on the given interval, substitute $x=-2$and $x=2$into the function $f\left(x\right)$.

So, substitute $x=-2$into the function localid="1647404834349" $\left(3\right)$.

The value of $f(-2)$is as follows:

$\begin{array}{rcl}f(-2)& =& \sqrt{4-{(-2)}^2}\\ & =& \sqrt{4-4}\\ & =& 0\end{array}$

Now, substitute $x=2$into the function $\left(3\right)$.

The value of $f\left(2\right)$is as follows:

$\begin{array}{rcl}f\left(2\right)& =& \sqrt{4-{\left(2\right)}^2}\\ & =& \sqrt{4-4}\\ & =& 0\end{array}$

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

localid="1647405225017" $\frac{∆y}{∆x}=\frac{f\left(2\right)-f(-2)}{-2-2}\dots \left(4\right)$

Substitute the value of $f\left(2\right)$and $f(-2)$into $\left(4\right)$.

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{0-0}{-2-2}\\ & =& \frac{0}{-4}\\ & =& 0\end{array}$

Thus, the average rate of change of the function $y=\sqrt{4-x^2}$is $0$.

Hence, from the average rate of change of $f\left(x\right)$conclude that ending point and starting point of the $y$-coordiante are the same.