Chapter 9.4, Problem 42E

To determine

To find: The graph for the given function.

Answer to Problem 42E

The required graph is shown in Figure 1.

Explanation of Solution

Given:

The given function is $g\left(x\right)=-\sin x-5$ .

Calculation:

Consider the given function is,

  $g\left(x\right)=-\sin x-5$

Compare the above equation with the general form $g\left(x\right)=a\cos \left(x-h\right)+k$ .

Then,

  $\begin{array}{l}a=\left|-1\right|\\ b=1\\ h=0\\ k=-5\end{array}$

From above and from the given function the period is obtained as,

  $\begin{array}{c}\frac{2\pi }{b}=\frac{2\pi }{1}\\ =2\pi \end{array}$

Then, determine the five key points as,

First,

  $\begin{array}{c}\left(0,0+k\right)=\left(0,0-5\right)\\ =\left(0,-5\right)\end{array}$

Second,

  $\begin{array}{c}\left(\frac{1}{2}\cdot \frac{2\pi }{b},0+k\right)=\left(\frac{1}{2}\cdot \frac{2\pi }{1},0-5\right)\\ =\left(\pi ,-5\right)\end{array}$

Third, the maximum is,

  $\begin{array}{c}\left(\frac{2\pi }{b},0+k\right)=\left(\frac{2\pi }{1},0-5\right)\\ =\left(2\pi ,-5\right)\end{array}$

Fourth point the maximum.

  $\begin{array}{c}\left(\frac{1}{4}\cdot \frac{2\pi }{b},a+k\right)=\left(\frac{1}{4}\cdot \frac{2\pi }{1},-1-5\right)\\ =\left(\frac{\pi }{2},-6\right)\end{array}$

Fifth point, the minimum.

  $\begin{array}{c}\left(\frac{3}{4}\cdot \frac{2\pi }{b},-a+k\right)=\left(\frac{3}{4}\cdot \frac{2\pi }{1},1-5\right)\\ =\left(\frac{3\pi }{2},-4\right)\end{array}$

From the points, the graph for the function is shown in Figure 1.

  

Figure 1