## Calculus with Applications (10th Edition)

$$y=f(x)=\frac{x+4}{x-1}\quad \text { from} \quad x=2\quad \text { to} \quad x=5$$ 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 $x= 2$: $$y^{\prime}(2)=\frac{-5}{((2)-1)^{2}}=\frac{-5}{(1)^{2}}=-5$$
$$y=f(x)=\frac{x+4}{x-1}\quad \text { from} \quad x=2\quad \text { to} \quad x=5$$ 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 $x= 2$: $$y^{\prime}(4)=\frac{-5}{((2)-1)^{2}}=\frac{-5}{(1)^{2}}=-5$$