Chapter 6.4, Problem 51AYU

To determine

To find: The graph of each function using method of key points. Also find domain and range.

Answer to Problem 51AYU

Solution:

Domain: $\left(-\infty , \infty \right)$

Range: $\left[-1.67, 1.67\right]$

Explanation of Solution

Given:

  $y=\frac{5}{3}\sin \left(-\frac{2\pi }{3}x\right)$

Calculation:

If $\omega >0$ , the amplitude and period of $y = A\sin \left(\omega x\right)$ and $y = A\cos \left(\omega x\right)$ are given by

Amplitude = $\left|A\right|$ Period $T=\frac{2\pi }{\omega }$ .

Since sine is an odd function, $y=\frac{5}{3}\sin \left(-\frac{2\pi }{3}x\right)=-\frac{5}{3}\sin \left(\frac{2\pi }{3}x\right)$

Comparing: $y=-\frac{5}{3}\sin \left(\frac{2\pi }{3}x\right)$ to $y = A\text{sin }\left(\omega x\right)$ , $A=-\frac{5}{3}$ and $\omega =\frac{2\pi }{3}$

Amplitude = $\left|A\right|=6$

Period $T=\frac{2\pi }{\omega }=\frac{2\pi }{\frac{2\pi }{3}}=3$

Because the amplitude is $\frac{5}{3}$ , the graph of $y=-\frac{5}{3}\sin \left(\frac{2\pi }{3}x\right)$ will lie between $-\frac{5}{3}$ and $\frac{5}{3}$ on the $y$ -axis. Because the period is $6$ , one cycle will begin at $x = 0$ and end at $x = 3$ .

Divide the interval $\left[0, 3\right]$ into four subintervals, each of length $\frac{3}{4}=\frac{3}{4}$ . The x-coordinates of the five key points are: $0, \frac{3}{4},\frac{3}{2}, \frac{9}{4},3$

To obtain the y-coordinates of the five key points of $y=-\frac{5}{3}\sin \left(\frac{2\pi }{3}x\right)$ , multiply the y-coordinates of the five key points of $y=\sin x$ by $A=-\frac{5}{3}.$ The five key points are

  $\left(0, 0\right) \left(\frac{3}{4},-\frac{5}{3}\right) \left(\frac{3}{2},0\right) \left(\frac{9}{4},\frac{5}{3}\right) \left(3, 0\right)$

  

Domain: $\left(-\infty , \infty \right)$

Range: $\left[-1.67, 1.67\right]$