Chapter 9.4, Problem 20E

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 20E

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

Explanation of Solution

Given:

The given function is $g\left(x\right)=\frac{1}{2}\cos 4\pi x$ .

Calculation:

Compare the function $g\left(x\right)=\frac{1}{2}\cos 4\pi x$ with $y=a\cos b\left(x-h\right)+k$ to find amplitude and period of the function.

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

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

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

Find the intercepts at $x=0,\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2}$ to draw the graph.

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

And,

  $\left(\frac{1}{4},\frac{1}{2}\cos \pi \right)=\left(\frac{1}{4},-\frac{1}{2}\right)$

Maximum point is $\left(0,\frac{1}{2}\right)$ and minimum point is $\left(\frac{1}{4},-\frac{1}{2}\right)$ .

Draw the graph of the function $g\left(x\right)=\frac{1}{2}\cos 4\pi x$ as follows.

  

Figure (1)

The graph of $g$ is a vertical shrink by a factor of $\frac{1}{2}$ and a horizontal shrink by a factor of $\frac{1}{4\pi }$ for the graph of $f$ .

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