Chapter 4, Problem 138RE

To determine

To find: an equation of a graph

Answer to Problem 138RE

An equation of a graph is

  $f(x)=4\cos \left(2x-\frac{3\pi }{2}\right)$

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

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

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:

Consider, the function,

  $f(x)=a\cos (bx-c)$

The maximum value is $+4$ and minimum value is $-4$

So, the $\text{Amplitude}=+4$

One cycle is completed for every $\pi$ , so that is the period

  $\begin{array}{l}\pi =\frac{2\pi }{b}\\ b=2\\ f(x)=4\cos \left(2x-c\right)\end{array}$

Find $c$ by setting the $x=\frac{3\pi }{4}$ and the angle is equal to $0$

  $\begin{array}{l}2x-c=0\\ 2\cdot \frac{3\pi }{4}-c=0\\ c=\frac{3\pi }{2}\end{array}$

The function is:

  $f(x)=4\cos \left(2x-\frac{3\pi }{2}\right)$

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

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

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:

Consider, the function,

  $f(x)=a\cos (bx-c)$

The maximum value is $+4$ and minimum value is $-4$

So, the $\text{Amplitude}=+4$

One cycle is completed for every $\pi$ , so that is the period

  $\begin{array}{l}\pi =\frac{2\pi }{b}\\ b=2\\ f(x)=4\cos \left(2x-c\right)\end{array}$

Find $c$ by setting the $x=\frac{3\pi }{4}$ and the angle is equal to $0$

  $\begin{array}{l}2x-c=0\\ 2\cdot \frac{3\pi }{4}-c=0\\ c=\frac{3\pi }{2}\end{array}$

The function is:

  $f(x)=4\cos \left(2x-\frac{3\pi }{2}\right)$