Chapter 2.1, Problem 82E

Solution

$\left(a\right)$

To calculate: The average rate of change of function $f\left(x\right)=x^2$ from $x=1$ to $x=3$ .

The average rate of function over the given interval is 4.

Given: $f\left(x\right)=x^2$ and the given interval is from $x=1$ to $x=3$ .

Formula used:

The average rate of change of function over the interval $a\le x\le b$ is given by the following expression:

  $\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

Find the rate of change of $f\left(x\right)=x^2$ over the interval $1\le x\le 3$ :

  $\begin{array}{l}f\left(1\right)=1^2=1\times 1=1\\ f\left(3\right)=3^2=3\times 3=9\end{array}$

The rate of change is calculated as follows:

  $\begin{array}{l}\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{9-1}{2}\\ \Rightarrow =4\end{array}$

$\left(b\right)$

To calculate: The average rate of change of function $f\left(x\right)=x^2$ from $x=2$ to $x=5$ .

The average rate of function over the given interval is $7$ .

Given: $f\left(x\right)=x^2$ and the given interval is from $x=2$ to $x=5$ .

Formula used:

The average rate of change of function over the interval $a\le x\le b$ is given by the following expression:

  $\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

Find the rate of change of $f\left(x\right)=x^2$ over the interval $2\le x\le 5$ :

  $\begin{array}{l}f\left(2\right)=2^2=2\times 2=4\\ f\left(5\right)=5^2=5\times 5=25\end{array}$

The rate of change is calculated as follows:

  $\begin{array}{l}\frac{f\left(5\right)-f\left(2\right)}{5-2}=\frac{25-4}{3}\\ \Rightarrow =7\end{array}$

$\left(c\right)$

To calculate: The average rate of change of function $f\left(x\right)=x^2$ from $x=a$ to $x=c$ .

The average rate of function over the given interval is $c+a$ .

Given: $f\left(x\right)=x^2$ and the given interval is from $x=a$ to $x=c$ .

Formula used:

The average rate of change of function over the interval $a\le x\le c$ is given by the following expression:

  $\frac{g\left(c\right)-g\left(a\right)}{c-a}$

Calculation:

Find the rate of change of $f\left(x\right)=x^2$ over the interval $a\le x\le c$ :

  $\begin{array}{l}f\left(a\right)=a^2\\ f\left(c\right)=c^2\end{array}$

The rate of change is calculated as follows:

  $\begin{array}{l}\frac{g\left(c\right)-g\left(a\right)}{c-a}=\frac{c^2-a^2}{c-a}\left[c^2-a^2=\left(c-a\right)\left(c+a\right)\right]\\ \Rightarrow =\frac{\left(c-a\right)\left(c+a\right)}{\left(c-a\right)}\\ \Rightarrow =\left(c+a\right)\end{array}$

$\left(d\right)$

To calculate: The average rate of change of function $g\left(x\right)=3x+2$ from $x=1$ to $x=3$ .

The average rate of function over the given interval is $3$ .

Given: $g\left(x\right)=3x+2$ and the given interval is from $x=1$ to $x=3$ .

Formula used:

The average rate of change of function over the interval $a\le x\le b$ is given by the following expression:

  $\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

Find the rate of change of $g\left(x\right)=3x+2$ over the interval $1\le x\le 3$ :

  $\begin{array}{l}g\left(1\right)=\left(3\times 1\right)+2\\ \Rightarrow =5\\ g\left(3\right)=\left(3\times 3\right)+2\\ \Rightarrow =11\end{array}$

The rate of change is calculated as follows:

  $\begin{array}{l}\frac{g\left(3\right)-g\left(1\right)}{3-1}=\frac{11-5}{2}\\ \Rightarrow =3\end{array}$

$\left(e\right)$

To calculate: The average rate of change of function $g\left(x\right)=3x+2$ from $x=1$ to $x=4$ .

The average rate of function over the given interval is $3$ .

Given: $g\left(x\right)=3x+2$ and the given interval is from $x=1$ to $x=4$ .

Formula used:

The average rate of change of function over the interval $a\le x\le b$ is given by the following expression:

  $\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

Find the rate of change of $g\left(x\right)=3x+2$ over the interval $1\le x\le 4$ :

  $\begin{array}{l}g\left(1\right)=\left(3\times 1\right)+2\\ \Rightarrow =5\\ g\left(5\right)=\left(3\times 4\right)+2\\ \Rightarrow =14\end{array}$

The rate of change is calculated as follows:

  $\begin{array}{l}\frac{g\left(4\right)-g\left(1\right)}{4-1}=\frac{14-5}{3}\\ \Rightarrow =3\end{array}$

$\left(f\right)$

To calculate: The average rate of change of function $g\left(x\right)=3x+2$ from $x=a$ to $x=c$ .

The average rate of function over the given interval is $3$ .

Given: $g\left(x\right)=3x+2$ and the given interval is from $x=a$ to $x=c$ .

Formula used:

The average rate of change of function over the interval $a\le x\le c$ is given by the following expression:

  $\frac{g\left(c\right)-g\left(a\right)}{c-a}$

Calculation:

Find the rate of change of $g\left(x\right)=3x+2$ over the interval $a\le x\le c$ :

  $\begin{array}{l}g\left(a\right)=\left(3\times a\right)+2\\ \Rightarrow =3a+2\\ g\left(c\right)=\left(3\times c\right)+2\\ \Rightarrow =3c+2\end{array}$

The rate of change is calculated as follows:

  $\begin{array}{l}\frac{g\left(c\right)-g\left(a\right)}{c-a}=\frac{3c+2-\left(3a+2\right)}{c-a}\\ \Rightarrow =\frac{3c+2-3a-2}{c-a}\\ \Rightarrow =\frac{3\left(c-a\right)}{c-a}\\ \Rightarrow =3\end{array}$

$\left(g\right)$

To calculate: The average rate of change of function $h\left(x\right)=7x-3$ from $x=a$ to $x=c$ .

The average rate of function over the given interval is $7$ .

Given: $h\left(x\right)=7x-3$ and the given interval is from $x=a$ to $x=c$ .

Formula used:

The average rate of change of function over the interval $a\le x\le c$ is given by the following expression:

  $\frac{h\left(c\right)-h\left(a\right)}{c-a}$

Calculation:

Find the rate of change of $h\left(x\right)=7x-3$ over the interval $a\le x\le c$ :

  $\begin{array}{l}h\left(a\right)=\left(7\times a-3\right)=7a-3\\ h\left(c\right)=\left(7\times c-3\right)=7c-3\end{array}$

The rate of change is calculated as follows:

  $\begin{array}{l}\frac{h\left(c\right)-h\left(a\right)}{c-a}=\frac{7c-3-\left(7a-3\right)}{c-a}\\ \Rightarrow =\frac{7c-3-7a+3}{c-a}\\ \Rightarrow =\frac{7\left(c-a\right)}{c-a}\end{array}$

On cancelling the like terms, the rate of change is $7$ .

  $\left(h\right)$

To calculate: The average rate of change of function $k\left(x\right)=mx+b$ from $x=a$ to $x=c$ .

The average rate of function over the given interval is $m$ .

Given: $k\left(x\right)=mx+b$ and the given interval is from $x=a$ to $x=c$ .

Formula used:

The average rate of change of function over the interval $a\le x\le c$ is given by the following expression:

  $\frac{k\left(c\right)-k\left(a\right)}{c-a}$

Calculation:

Find the rate of change of $k\left(x\right)=mx+b$ over the interval $a\le x\le c$ :

  $\begin{array}{l}k\left(a\right)=\left(m\times a\right)+b=ma+b\\ k\left(c\right)=\left(m\times c\right)+b=mc+b\end{array}$

The rate of change is calculated as follows:

  $\begin{array}{l}\frac{k\left(c\right)-k\left(a\right)}{c-a}=\frac{mc+b-\left(ma-b\right)}{c-a}\\ \Rightarrow =\frac{m\left(c-a\right)}{c-a}\\ \Rightarrow =m\end{array}$

$\left(i\right)$

To calculate: The average rate of change of function $l\left(x\right)=x^3$ from $x=a$ to $x=c$ .

The average rate of function over the given interval is $c^2+a^2+ac$ .

Given: $l\left(x\right)=x^3$ and the given interval is from $x=a$ to $x=c$ .

Formula used:

The average rate of change of function over the interval $a\le x\le c$ is given by the following expression:

  $\frac{l\left(c\right)-l\left(a\right)}{c-a}$

Calculation:

Find the rate of change of $l\left(x\right)=x^3$ over the interval $a\le x\le c$ :

  $\begin{array}{l}l\left(a\right)=a^3\\ l\left(c\right)=c^3\end{array}$

The rate of change is calculated as follows:

  $\begin{array}{l}\frac{l\left(c\right)-l\left(a\right)}{c-a}=\frac{c^3-a^3}{c-a}\left[c^3-a^3=\left(c-a\right)\left(c^2+a^2+ab\right)\right]\\ \Rightarrow =\frac{\left(c-a\right)\left(c^2+a^2+ab\right)}{c-a}\\ \Rightarrow =\left(c^2+a^2+ab\right)\end{array}$