Chapter 2.4, Problem 4E

(a)

To determine

To find: The average rate of change of the function $f\left(x\right)=\ln x$ over the interval $\left[1,4\right]$.

Answer to Problem 4E

The average rate of change of the function $f\left(x\right)=\ln x$ over the interval $\left[1,4\right]$ is equal to $0.46$ approximately.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\ln x$ and the interval is $\left[1,4\right]$.

Calculation:

The average rate of change of function $f\left(x\right)$ over the interval $\left[a,b\right]$ is $\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

The formula for the average rate of change of function $f\left(x\right)=\ln x$ over the interval $\left[1,4\right]$ is:

  $\frac{f\left(4\right)-f\left(1\right)}{4-1}=\frac{f\left(4\right)-f\left(1\right)}{3}$

The function is given as $f\left(x\right)=\ln x$.

Substitute $4$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=\ln x\\ f\left(4\right)=\ln 4\\ f\left(4\right)\approx 1.39\end{array}$

Substitute $1$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=\ln x\\ f\left(1\right)=\ln 1\\ f\left(1\right)=0\end{array}$

Substitute $1.39$ for $f\left(4\right)$ and $0$ for $f\left(1\right)$ in the formula.

  $\begin{array}{c}\frac{f\left(4\right)-f\left(1\right)}{3}=\frac{1.39-0}{3}\\ =\frac{1.39}{3}\\ \approx 0.46\end{array}$

Therefore, the average rate of change of the function $f\left(x\right)=\ln x$ over the interval $\left[1,4\right]$ is $0.46$ approximately.

(b)

To determine

To find: The average rate of change of the function $f\left(x\right)=\ln x$ over the interval $\left[100,103\right]$.

Answer to Problem 4E

The average rate of change of the function $f\left(x\right)=\ln x$ over the interval $\left[100,103\right]$ is $0.01$.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\ln x$ and the interval is $\left[100,103\right]$.

Calculation:

The average rate of change of function $f\left(x\right)$ over the interval $\left[a,b\right]$ is $\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

The formula for the average rate of change of function $f\left(x\right)=\ln x$ over the interval $\left[100,103\right]$ is:

  $\frac{f\left(103\right)-f\left(100\right)}{103-100}=\frac{f\left(103\right)-f\left(100\right)}{3}$

The function is given as $f\left(x\right)=\ln x$.

Substitute $103$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=\ln x\\ f\left(103\right)=\ln \left(103\right)\\ f\left(103\right)\approx 4.63\end{array}$

Substitute $100$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=\ln x\\ f\left(100\right)=\ln \left(100\right)\\ f\left(100\right)\approx 4.60\end{array}$

Substitute $4.63$ for $f\left(103\right)$ and $4.60$ for $f\left(100\right)$ in the formula.

  $\begin{array}{c}\frac{f\left(103\right)-f\left(100\right)}{3}=\frac{4.63-4.60}{3}\\ =\frac{0.03}{3}\\ =0.01\end{array}$

Therefore, the average rate of change of the function $f\left(x\right)=\ln x$ over the interval $\left[100,103\right]$ is $0.01$ approximately.