Chapter 4, Problem 136RE

To determine

To Find: An equation of a graph

Answer to Problem 136RE

  $Anequationofagraphis$

  $f(x)=\frac{1}{3}\cos (\pi x)$

Explanation of Solution

Given information: graph

  

Concept used: The graph of $f(x)=a\sin (bx-c)$ And $f(x)=a\cos (bx-c)$

have the following characteristics

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

The left and right end points of a one-cycle interval can be determined by solving the equation

  $bx-c=0$ and $bx-c=2\pi$

Calculation:

  $f(x)=a\cos (bx-c)$

The maximum value is $+\frac{1}{2}$

and minimum value is $-\frac{1}{2}$

So, the $Amplitude=+\frac{1}{2}$

One cycle is completed for every $2$ , So that is the period

  $\begin{array}{l}2=\frac{2\pi }{b}\\ b=\pi \end{array}$

  $f(x)=\frac{1}{3}\cos (\pi x-c)$

Find $c$

the graph starts at the maximum and goes down

so, $c=0$

The function is

  $f(x)=\frac{1}{3}\cos (\pi x)$