Chapter 9.4, Problem 19E

To determine

To identify: The amplitude and period of the function then graph the function and describe the function of $g$ as a transformation of the graph of its parent function.

Answer to Problem 19E

The amplitude is $\frac{1}{3}$ and period is $\frac{\pi }{2}$ and graph is shown in figure (1).

Explanation of Solution

Given:

The given function is $g\left(x\right)=\frac{1}{3}\cos 4x$ .

Calculation:

Compare the function $g\left(x\right)=\frac{1}{3}\cos 4x$ with $y=a\cos b\left(x-h\right)+k$ to find amplitude and period of the function.

  $\begin{array}{c}a=\frac{1}{3}\\ \text{Period}=\frac{2\pi }{b}\\ =\frac{2\pi }{4}\\ =\frac{\pi }{2}\end{array}$

The horizontal shift is $h=0$ and vertical shifts is $k=0$ .

Draw the midline of the graph, $y=0$ .

Find the intercepts at $x=0,\frac{\pi }{8},\frac{\pi }{4},\frac{3\pi }{8},\frac{\pi }{2}$ to draw the graph.

  $\begin{array}{c}\left(0,\frac{1}{3}\cos 0\right)=\left(0,\frac{1}{3}\right)\\ \left(\frac{\pi }{8},\frac{1}{3}\cos \frac{\pi }{2}\right)=\left(\frac{\pi }{8},0\right)\\ \left(\frac{3\pi }{8},\frac{1}{3}\cos \frac{3\pi }{2}\right)=\left(\frac{3\pi }{8},0\right)\\ \left(\frac{\pi }{2},\frac{1}{3}\cos 2\pi \right)=\left(\frac{\pi }{2},\frac{1}{3}\right)\end{array}$

And,

  $\left(\frac{\pi }{4},\frac{1}{3}\cos \pi \right)=\left(\frac{\pi }{4},-\frac{1}{3}\right)$

Maximum point is $\left(0,\frac{1}{3}\right)$ and minimum point is $\left(\frac{\pi }{4},-\frac{1}{3}\right)$ .

Draw the graph of the function $g\left(x\right)=\frac{1}{3}\cos 4x$ as follows.

  

Figure (1)

The graph of $g$ is vertical shrink by $\frac{1}{3}$ , and horizontal shrink by $\frac{1}{4}$ of the graph of $f$ .

Therefore, the amplitude is $\frac{1}{3}$ and period is $\frac{\pi }{2}$ and graph is shown in figure (1).