Chapter 8, Problem 4CA

To determine

Find the order of the functions from the least average rate of change to the greatest average rate of change on the given interval.

Answer to Problem 4CA

The order is $B,A,C$ and $D$ .

Explanation of Solution

Given:

The given functions are

A $f\left(x\right)=4\sqrt{x+2}$

B $x$ and $y$ vary inversely as each other, when $x=5$ and $y=2$ .

C

  

D

  

The given interval is $1\le x\le 4$ .

Calculation:

A $f\left(x\right)=4\sqrt{x+2}$ , $1\le x\le 4$

  $\begin{array}{l}f\left(1\right)=4\sqrt{1+2}=4\sqrt{3}=6.928\\ f\left(4\right)=4\sqrt{4+2}=4\sqrt{6}=9.798\end{array}$

The average rate of change $=\frac{y_2-y_1}{x_2-x_1}$

  $\begin{array}{l}{x}_1=1,y_1=6.928\\ x_2=4,y_2=9.798\end{array}$

  $\frac{y_2-y_1}{x_2-x_1}=\frac{9.798-6.928}{4-1}=\frac{2.87}{3}\approx 0.957$

B $x$ and $y$ vary inversely as each other, when $x=5$ and $y=2$ . $1\le x\le 4$

  $\begin{array}{l}x=\frac{k}{y}\\ \Rightarrow 5=\frac{k}{2}\\ \Rightarrow k=10\\ xy=10\end{array}$

If $x=1$ , $y=\frac{10}{1}=10$

If $x=4$ , $y=\frac{10}{4}=2.5$

The average rate of change $=\frac{y_2-y_1}{x_2-x_1}$

  $\begin{array}{l}{x}_1=1,y_1=10\\ x_2=4,y_2=2.5\end{array}$

  $\frac{y_2-y_1}{x_2-x_1}=\frac{2.5-10}{4-1}=-\frac{7.5}{3}=-2.5$

C

  

From the graph

  $\begin{array}{l}{x}_1=1,y_1=2\\ x_2=4,y_2=11\end{array}$

The average rate of change $=\frac{y_2-y_1}{x_2-x_1}$

  $\frac{y_2-y_1}{x_2-x_1}=\frac{11-2}{4-1}=\frac{9}{3}=3$

D

  

From the table

  $\begin{array}{l}{x}_1=1,y_1=-4\\ x_2=4,y_2=5\end{array}$

The average rate of change $=\frac{y_2-y_1}{x_2-x_1}$

  $\frac{y_2-y_1}{x_2-x_1}=\frac{5-\left(-4\right)}{4-1}=\frac{9}{3}=3$

Hence the order is $B,A,C$ and $D$ .