91Ó°ÊÓ

We have to draw a sequence of graphs that illustrates how to do this for z < 2, and then make specific calculations for z = 1, z = 1.5, z = 1.75, and z = 1.9.

Secant lines for the function is :

For x=2, z=1, the value of $f\left(2\right)=4,f\left(1\right)=1.25$

Therefore, the average rate of change is :

$\begin{array}{r}\frac{f\left(1\right)−f\left(2\right)}{1−2}=\frac{1.25−4}{−1}\\ =\frac{−2.75}{−1}\\ =2.75\end{array}$

Here the value of h is :

$f\left(2\right)=4,f\left(1.5\right)=3.375$

Therefore the average rate of change is :

$\begin{array}{r}\frac{f\left(1.5\right)−f\left(2\right)}{1.5−2}=\frac{3.375−4}{1}\\ =\frac{−0.625}{−0.5}\\ =1.25\end{array}$

Here, the value of h is :$z−x=−0.5$

For x=2, z=1, the value of$f\left(2\right)=4,f\left(1.75\right)=3.719$

Therefore, the average rate of change is :

$\begin{array}{r}\frac{f\left(1.75\right)−f\left(2\right)}{1.75−2}=\frac{3.718−4}{−0.25}\\ =\frac{−0.282}{−0.25}\\ =1.25\end{array}$

Here, the value of h is : $z−x=−0.25$

For x=2, z=1.9, the value of $f\left(2\right)=4,f\left(1.9\right)=3.895$

Therefore the average rate of change is :

$\begin{array}{r}\frac{f\left(1.9\right)−f\left(2\right)}{1.9−2}=\frac{3.895−4}{−0.1}\\ =\frac{−0.105}{−0.1}\\ =1.05\end{array}$

Thus,

$h=-0.1$