Chapter 1.6, Problem 20E

(a)

To determine

To find: The period of the function $y=5\sin \frac{x}{2}$ .

Answer to Problem 20E

The period of the function is $4\pi$ .

Explanation of Solution

Given information:The given function is.

Calculation:

The period of a function $y=A\sin \left(Bx+C\right)+D$ is given by $\frac{2\pi }{B}$ .

Consider the given function $y=5\sin \frac{x}{2}$ .

So, the period of the given function is:

  $\begin{array}{c}\frac{2\pi }{B}=\frac{1}{2}\\ B=4\pi \end{array}$

Therefore, the period of the function is $4\pi$ .

(b)

To determine

To find: The amplitude of the function $y=5\sin \frac{x}{2}$ .

Answer to Problem 20E

The amplitude of the function is $5$ .

Explanation of Solution

Given information:The given function is $y=5\sin \frac{x}{2}$ .

Calculation:

The general form of sine function is:

  $y=A\sin \left(Bx+C\right)+D$

Here, $\left|A\right|$ is the amplitude.

The amplitude of the given function $y=5\sin \frac{x}{2}$ is:

  $\left|A\right|=5$

Therefore, theamplitude of the function is $5$ .

(c)

To determine

To find: The viewing window shown in the graph given.

Answer to Problem 20E

The viewing window for the given graph is $\left[-4\pi ,4\pi \right]$ by $\left[-2,2\right]$ .

Explanation of Solution

Given information:The given function is $y=5\sin \frac{x}{2}$

and the given graph is:

  

Calculation:

The function $y=5\sin \frac{x}{2}$ has a period of $\pi$ and it repeats its values twice to the right and twice to the left of origin in the given graph.

So, the value of $x$ lies between $-4\pi$ and $4\pi$ . As shown in graph the values of $y$ lies between $-2$ and $2$ .

Therefore, the viewing window for the given graph is $\left[-4\pi ,4\pi \right]$ by $\left[-2,2\right]$ .