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

  $\begin{array}{l}f\left(x\right)=\cos x\\ g\left(x\right)=\cos \left(x+\pi \right)\end{array}$

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

  $a=1$ and $b=1$

Thus,

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

And

  $\begin{array}{c}\text{Period=}\frac{2\pi }{b}\\ =\frac{2\pi }{\left(1\right)}\\ =2\pi \end{array}$

The interval $\left[0,2\pi \right]$ corresponds to one cycle on the graph.

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

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

The left and right endpoints of one cycle interval can be determined by solving the equations

  $bx-c=0$ and $bx-c=2\pi$

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

  $a=1$ and $b=1$

Thus,

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

And

  $\begin{array}{c}\text{Period=}\frac{2\pi }{b}\\ =\frac{2\pi }{\left(1\right)}\\ =2\pi \end{array}$

The left end point is,

  $\begin{array}{c}x+\pi =0\\ x=-\pi \end{array}$

The right end point is,

  $\begin{array}{c}x+\pi =2\pi \\ x=\pi \end{array}$

So interval $\left[-\pi ,\pi \right]$ corresponds to one cycle on the graph.

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 $f\left(x\right)\pi$ units to the left.