Answer
$$
y=f(x)=\frac{x+4}{x-1}\quad \text { from} \quad x=2\quad \text { to} \quad x=5
$$
The average rate of change of $ f(x) $ with respect to $x$ as $x$ changes from $a=2 $ to $ b=5$ is
$$
\begin{split}
\text {Average rate of change}
&=\frac{f(b)-f(a)}{b-a}\\
& =\frac{f(5)-f(2)}{5-2}\\
& =\frac{(\frac{(5)+4}{(5)-1})-(\frac{(2)+4}{(2)-1})}{5-2}\\
& =\frac{(\frac{9}{4})-(6)}{3}\\
& =\frac{(\frac{-15}{4})}{3}\\
& =\frac{-5}{4}\\
\end{split}
$$
and we can find that :
$$
y^{\prime}(x)=\frac{(x-1)(1)-(x+4)(1)}{(x-1)^{2}}=\frac{-5}{(x-1)^{2}}
$$
Instantaneous rate of change at $x= 2$:
$$
y^{\prime}(2)=\frac{-5}{((2)-1)^{2}}=\frac{-5}{(1)^{2}}=-5
$$
Work Step by Step
$$
y=f(x)=\frac{x+4}{x-1}\quad \text { from} \quad x=2\quad \text { to} \quad x=5
$$
The average rate of change of $ f(x) $ with respect to $x$ as $x$ changes from $a=2 $ to $ b=5$ is
$$
\begin{split}
\text {Average rate of change}
&=\frac{f(b)-f(a)}{b-a}\\
& =\frac{f(5)-f(2)}{5-2}\\
& =\frac{(\frac{(5)+4}{(5)-1})-(\frac{(2)+4}{(2)-1})}{5-2}\\
& =\frac{(\frac{9}{4})-(6)}{3}\\
& =\frac{(\frac{-15}{4})}{3}\\
& =\frac{-5}{4}\\
\end{split}
$$
and we can find that :
$$
y^{\prime}(x)=\frac{(x-1)(1)-(x+4)(1)}{(x-1)^{2}}=\frac{-5}{(x-1)^{2}}
$$
Instantaneous rate of change at $x= 2$:
$$
y^{\prime}(4)=\frac{-5}{((2)-1)^{2}}=\frac{-5}{(1)^{2}}=-5
$$
