Chapter 2.4, Problem 5E

(a)

To determine

To find: The average rate of change of the function $f\left(x\right)=\cot x$ over the interval $\left[\frac{\pi }{4},\frac{3\pi }{4}\right]$.

Answer to Problem 5E

The average rate of change of the function $f\left(x\right)=\cot x$ over the interval $\left[\frac{\pi }{4},\frac{3\pi }{4}\right]$ is $-1.273$ approximately.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\cot x$ and the interval is $\left[\frac{\pi }{4},\frac{3\pi }{4}\right]$.

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 average rate of change of function $f\left(x\right)=\cot x$ over the interval $\left[\frac{\pi }{4},\frac{3\pi }{4}\right]$ is:

  $\frac{f\left(\frac{3\pi }{4}\right)-f\left(\frac{\pi }{4}\right)}{\frac{3\pi }{4}-\frac{\pi }{4}}=\frac{2\left(f\left(\frac{3\pi }{4}\right)-f\left(\frac{\pi }{4}\right)\right)}{\pi }$

The function is given as $f\left(x\right)=\cot x$ . Substitute $\frac{3\pi }{4}$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=\cot x\\ f\left(\frac{3\pi }{4}\right)=\cot \left(\frac{3\pi }{4}\right)\\ f\left(\frac{3\pi }{4}\right)=-1\end{array}$

Substitute $\frac{\pi }{4}$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=\cot x\\ f\left(\frac{\pi }{4}\right)=\cot \left(\frac{\pi }{4}\right)\\ f\left(\frac{\pi }{4}\right)=1\end{array}$

Substitute $-1$ for $f\left(\frac{3\pi }{4}\right)$ and $1$ for $f\left(\frac{\pi }{4}\right)$ in the formula.

  $\begin{array}{c}\frac{2\left(f\left(\frac{3\pi }{4}\right)-f\left(\frac{\pi }{4}\right)\right)}{\pi }=\frac{2\left(-1-1\right)}{\pi }\\ =\frac{2\times -2}{\pi }\\ =\frac{-4}{\pi }\\ \approx -1.273\end{array}$

Therefore, the average rate of change of the function $f\left(x\right)=\cot x$ over the interval $\left[\frac{\pi }{4},\frac{3\pi }{4}\right]$ is $-1.273$ approximately.

(b)

To determine

To find: The average rate of change of the function $f\left(x\right)=\cot x$ over the interval $\left[\frac{\pi }{6},\frac{\pi }{2}\right]$.

Answer to Problem 5E

The average rate of change of the function $f\left(x\right)=\cot x$ over the interval $\left[\frac{\pi }{6},\frac{\pi }{2}\right]$ is $-1.65$.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\cot x$ and the interval is $\left[\frac{\pi }{6},\frac{\pi }{2}\right]$.

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 $f\left(x\right)=\cot x$ over the interval $\left[\frac{\pi }{6},\frac{\pi }{2}\right]$ is:

  $\frac{f\left(\frac{\pi }{2}\right)-f\left(\frac{\pi }{6}\right)}{\frac{\pi }{2}-\frac{\pi }{6}}=\frac{3\left(f\left(\frac{\pi }{2}\right)-f\left(\frac{\pi }{6}\right)\right)}{\pi }$

The function is given as $f\left(x\right)=\cot x$ . Substitute $\frac{\pi }{2}$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=\cot x\\ f\left(\frac{\pi }{2}\right)=\cot \left(\frac{\pi }{2}\right)\\ f\left(\frac{\pi }{2}\right)=0\end{array}$

Substitute $\frac{\pi }{6}$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=\cot x\\ f\left(\frac{\pi }{6}\right)=\cot \left(\frac{\pi }{6}\right)\\ f\left(\frac{\pi }{6}\right)=\sqrt{3}\\ f\left(\frac{\pi }{6}\right)\approx 1.73\end{array}$

Substitute $0$ for $f\left(\frac{\pi }{2}\right)$ and $1.73$ for $f\left(\frac{\pi }{6}\right)$ in the formula.

  $\begin{array}{c}\frac{3\left(f\left(\frac{\pi }{2}\right)-f\left(\frac{\pi }{6}\right)\right)}{\pi }=\frac{3\left(0-1.73\right)}{\pi }\\ =\frac{3\times -1.73}{\pi }\\ =-\frac{5.19}{\pi }\\ \approx -1.65\end{array}$

Therefore, the average rate of change of the function $f\left(x\right)=\cot x$ over the interval $\left[\frac{\pi }{6},\frac{\pi }{2}\right]$ is $-1.65$ approximately.