Chapter 0.6, Problem 22E

a.

To determine

To specify: the period that is shown in the given graph of $y=\cos \left(\pi x\right)$ .

Answer to Problem 22E

The period that is shown in the given graph of $y=\cos \left(\pi x\right)$ is 2.

Explanation of Solution

Given information:

The graph:

  

Formula Used:

The period of the function $y=A\cos \left(Bx+C\right)+D$ , where A, B, C, and D are real constants is $\frac{2\pi }{B}$ .

By the above description, the period of $y=A\cos \left(Bx+C\right)+D$ , where A, B, C, and D are real constants is $\frac{2\pi }{B}$ .

Here, in this case the given function is $y=\cos \left(\pi x\right)$ , so the period is:

  $\frac{2\pi }{B}=\frac{2\pi }{\pi }=2$

b.

To determine

To specify: the amplitude that is shown in the given graph of $y=\cos \left(\pi x\right)$ .

Answer to Problem 22E

The amplitude that is shown in the given graph of $y=\cos \left(\pi x\right)$ is $1$ .

Explanation of Solution

Given information:

The graph:

  

Formula Used:

The amplitude of the function $y=A\cos \left(Bx+C\right)+D$ , where A, B, C, and D are real constants is $\left|A\right|$ .

By the above description, the amplitude of $y=A\cos \left(Bx+C\right)+D$ , where A, B, C, and D are real constants is $\left|A\right|$ .

Here, in this case the given function is $y=\cos \left(\pi x\right)$ which can be written as $y=1\cos \left(\pi x\right)$ , so the amplitude is:

  $\left|A\right|=\left|1\right|=1$

c.

To determine

To identify: the viewing window that is shown in the given graph of $y=\cos \left(\pi x\right)$ .

Answer to Problem 22E

  $\left[-4,4\right]\text{ by }\left[-2,2\right]$

Explanation of Solution

Given information:

The graph:

  

By looking at the graph, it can be observed that the graph shows 4 complete periods. Also from part (a), the period of this function is 2.

Thus, the horizontal window must be $\left[-4,4\right]$ .

Also, from part (b), it is known that the amplitude of this function is 1. This means that the maximum value of this function is 1 and the minimum value is -1.

Now, looking at the graph it can be observed that there is one unit above and one unit below the graph of the function.

So, the vertical window is $\left[-2,2\right]$ .

Hence, the viewing window is $\left[-4,4\right]\text{ by }\left[-2,2\right]$ .