Chapter 2, Problem 83EP

(a).

To determine

To graph: The function $f\left(x\right)=\frac{\cos \left(x\right)}{\cos \left(x\right)-2}\text{for}-2\pi \le x\le 2\pi$ .

Answer to Problem 83EP

  

Explanation of Solution

Given:

  $f\left(x\right)=\frac{\cos \left(x\right)}{\cos \left(x\right)-2}\text{for}-2\pi \le x\le 2\pi$

Concept Used:

The graph of the function using graphing calculator is:

  

b.

To determine

To calculate: The derivative of $f\left(x\right)=\frac{\cos \left(x\right)}{\cos \left(x\right)-2}\text{for}-2\pi \le x\le 2\pi$ .

Answer to Problem 83EP

Derivative of $f\text{'}\left(x\right)=$

  $\frac{2\sin \left(x\right)}{{\left(\cos \left(x\right)-2\right)}^2}$ .

Explanation of Solution

Given:

  $f\left(x\right)=\frac{\cos \left(x\right)}{\cos \left(x\right)-2}\text{for}-2\pi \le x\le 2\pi$

Concept Used:

1. $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{\frac{d}{dx}\left[u\left(x\right)\right]\cdot v(x)-u(x)\cdot \frac{d}{dx}\left[v(x)\right]}{v{(x)}^2}$

Where $u(x),v(x)$ are functions of $x$ .

2. $\frac{d}{dx}\left[\cos \left(x\right)\right]=-\sin \left(x\right)$

Calculation:

  $\begin{array}{c}\frac{d}{dx}\left[f(x)\right]=\frac{d}{dx}\left[\frac{\cos \left(x\right)}{\cos \left(x\right)-2}\right]\\ =\frac{\frac{d}{dx}[\cos (x)]\cdot (\cos (x)-2)-\cos (x)\cdot \frac{d}{dx}[\cos (x)-2]}{{(\cos (x)-2)}^2}\\ =\frac{(-\sin (x))(\cos (x)-2)-\cos (x)(\frac{d}{dx}[\cos (x)]+\frac{d}{dx}[-2])}{{(\cos (x)-2)}^2}\\ =\frac{-(\cos (x)-2)\sin (x)-\cos (x)((-\sin (x))+0)}{{(\cos (x)-2)}^2}\\ =\frac{\cos \left(x\right)\sin \left(x\right)-\left(\cos \left(x\right)-2\right)\sin \left(x\right)}{{\left(\cos \left(x\right)-2\right)}^2}\\ =\frac{\cos \left(x\right)\sin \left(x\right)}{{\left(\cos \left(x\right)-2\right)}^2}-\frac{\sin \left(x\right)}{\cos \left(x\right)-2}\\ =\frac{2\sin \left(x\right)}{{\left(\cos \left(x\right)-2\right)}^2}\end{array}$

Conclusion:

Derivative of $f\text{'}\left(x\right)=$

  $\frac{2\sin \left(x\right)}{{\left(\cos \left(x\right)-2\right)}^2}$ .

c.

To determine

To calculate: All the values in the domain of f for which $f\text{'}(x)=0$ .

Answer to Problem 83EP

  $-2\pi ,-\pi ,0,\pi ,2\pi$ .

Explanation of Solution

Given:

  $f\left(x\right)=\frac{\cos \left(x\right)}{\cos \left(x\right)-2}\text{for}-2\pi \le x\le 2\pi$ .

Concept Used:

Solution can be found from the graph. $f\text{'}(x)=0$ where the graph of $f(x)$ has slope value of zero.

Calculation:

For $f\text{'}(x)=0$ , the numerator term $-2\sin x$ has to be zero and the denominator should be zero.

  $\begin{array}{c}-2\sin \left(x\right)=0\\ \sin \left(x\right)=0\\ x=\pm n\pi \end{array}$

Apart from the numerical solution, solution can be found from the graph $f\text{'}(x)=0$ where the graph of $f(x)$ has slope value of zero.

Conclusion:

All the values in the domain of f for which $f\text{'}(x)=0$ are $-2\pi ,-\pi ,0,\pi ,2\pi$

d.

To determine

The range of $f(x)$ using the information gathered from solution of parts (a) and (c).

Answer to Problem 83EP

The range of $f(x)$ is $\left[-1,\frac{1}{3}\right]$

Explanation of Solution

Given:

  $f\left(x\right)=\frac{\cos \left(x\right)}{\cos \left(x\right)-2}\text{for}-2\pi \le x\le 2\pi$

Concept Used:

Calculation:

From solution of part (c) it is found that $f(x)$ has extreme points (maxima or minima) along the y-axis when $x=$ $\pm n\pi$ , since $f\text{'}(x)=0$ or it can be said that the slope is zero for $f(x)$ at these points (that means these x values correspond to the minima or maxima of $f(x)$ ) From the graph of part (a), it is seen that $f(x)$ has maximum value when n is odd and has minimum value when n is even. Also, the graph shows f(x) is period with period of $2\pi$ .

Minimum value of $f\left(x\right)\left(whenx=0\right)=-1$

Maximum value of $f\left(x\right)\left(whenx=\pi \right)=\frac{1}{3}$

Conclusion:

The range of $f(x)$ is $\left[-1,\frac{1}{3}\right]$ .