Chapter 11, Problem 32RE

Solution

a)

To find:the average rate of change of the function over a given interval.

The average rate of change of $y$ with respect to $x$ on the interval $⌊3,3.01⌋$ is $-0.12$ .

Given information:

The expression $f\left(x\right)=\frac{3}{x+2}$ over the interval $⌊3,3.01⌋$ .

Formula used:

If $y=f\left(x\right)$ then the average rate of change of $y$ with respect to $x$ on the interval $\left[a,b\right]$ is $\frac{\Delta y}{\Delta x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ .

Calculation:

Compute the average rate:

  $\begin{array}{c}\frac{\Delta y}{\Delta x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ =\frac{f\left(3.01\right)-f\left(3\right)}{3.01-3}\\ =\frac{\left(\frac{3}{3.01+2}\right)-\left(\frac{3}{3+2}\right)}{0.01}\\ =\frac{\left(\frac{3}{5.01}\right)-\left(\frac{3}{5}\right)}{0.01}\\ =\frac{0.5988-0.6}{0.01}\\ =-\frac{0.0012}{0.01}\\ =-0.12\end{array}$

The average rate of change of $y$ with respect to $x$ on the interval $⌊3,3.01⌋$ is $-0.12$ .

b)

To find:the instantaneous rate of change at a given point.

The instantaneous rate of change of $f\left(x\right)=\frac{3}{x+2}$ at $x=3$ is $-\frac{3}{25}$ .

Given information:

The expression $f\left(x\right)=\frac{3}{x+2}$ at $x=3$ .

Formula used:

The instantaneous rate of change of $f$ at $x=a$ is given as $f^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$

Calculation:

Compute the limit:

  $\begin{array}{c}{f}^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}\\ f^{\prime }\left(3\right)=\underset{x\to 3}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3}\\ =\underset{x\to 3}{\lim }\frac{\left(\frac{3}{x+2}\right)-\left(\frac{3}{3+2}\right)}{x-3}\\ =\underset{x\to 3}{\lim }\frac{\left(\frac{3}{x+2}\right)-\left(\frac{3}{5}\right)}{x-3}\end{array}$

Further, simplify.

  $\begin{array}{c}{f}^{\prime }\left(3\right)=\underset{x\to 3}{\lim }\frac{\left(15-3x-6\right)/5\left(x+2\right)}{x-3}\\ =\underset{x\to 3}{\lim }\frac{-3\left(x-3\right)}{5\left(x-3\right)\left(x+2\right)}\\ =\underset{x\to 3}{\lim }\frac{-3}{5\left(x+2\right)}\\ =\frac{-3}{5\left(\left(3\right)+2\right)}\\ =-\frac{3}{25}\end{array}$

Thus, the instantaneous rate of change of $f\left(x\right)=\frac{3}{x+2}$ at $x=3$ is $-\frac{3}{25}$ .