Chapter 4.5, Problem 20E

To determine

To describe: the relationship between the graphs of $f$ and $g$ .

Explanation of Solution

Given information:

Given functions are

  $\begin{array}{l}f\left(x\right)=\cos 4x\\ g\left(x\right)=-2+xos4x\end{array}$

Concept used:

For the standard equation is $y=a\cos bx$ .

Amplitude is $\left|a\right|$ and period is $\frac{2\pi }{b}$ .

Calculation:

Comparing $f\left(x\right)=\cos 4x$ with the equation $y=a\cos bx$ ,

  $a=1$ and $b=4$

Thus,

  $\begin{array}{c}\text{Amplitude=}\left|1\right|\\ =1\end{array}$

And

  $\begin{array}{c}\text{Period=}\frac{2\pi }{b}\\ =\frac{2\pi }{\left(4\right)}\\ =\frac{\pi }{2}\end{array}$

The amplitude of $y=d+a\sin \left(bx-c\right)$ and $y=d+a\cos \left(bx-c\right)$ have following characteristics.

  $\begin{array}{c}\text{Amplitude=}\left|a\right|\\ \text{Period=}\frac{2\pi }{b}\end{array}$

The shift is $d$ units upward for $d>0$ and $d$ units downward for $d<0$

Comparing $g\left(x\right)=-2+\cos 4x$ with the equation $y=a\cos bx$ ,

  $a=1$ and $b=4$

Thus,

  $\begin{array}{c}\text{Amplitude=}\left|1\right|\\ =1\end{array}$

And

  $\begin{array}{c}\text{Period=}\frac{2\pi }{b}\\ =\frac{2\pi }{\left(4\right)}\\ =\frac{\pi }{2}\end{array}$

Here $d=-2$ , so the graph will be shifted 2 unit downwards from the graph of $f\left(x\right)$ .

Now plot the graphs of $f\left(x\right)$ and $g\left(x\right)$ :

  

By above graph it is seen that $g\left(x\right)$ is a shift of $f\left(x\right)$ two units downward.