Chapter 4.4, Problem 38E

Solution

To identify: The maximum and minimum values and the zeros of the function $y=\frac{1}{2}\sin x$ in the interval $\left[-2\pi ,2\pi \right]$ .

The function has maximum $\frac{1}{2}$ at $-\frac{3\pi }{2}\text{ and }\frac{\pi }{2}$ , minimum $-\frac{1}{2}$ at $-\frac{\pi }{2}\text{ and }\frac{3\pi }{2}$ , and zeros at $0,\pm \pi ,\pm 2\pi$ .

Given information:

The given function is $y=\frac{1}{2}\sin x$ .

Explanation:

Consider the given function: $y=\frac{1}{2}\sin x$

Compare the function $y=\frac{1}{2}\sin x$ with the standard function $y=a\sin bx$ .

Then the values are $a=\frac{1}{2}\text{ and }b=1$ .

Here $a$ is the amplitude of the function and its value is 2.

Now find the period of the function.

  $\text{Period}=\frac{2\pi }{\left|b\right|}$

Substitute the values $b=1$ in the formula:

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

Thus the graph of $y=\frac{1}{2}\sin x$ is a vertical shrink of the graph of $y=\sin x$ by a factor of $\frac{1}{2}$ and there is no horizontal stretch or shrink.

The characteristics of the function $y=\sin x$

:

Maximum is 1 at $-\frac{3\pi }{2}\text{ and }\frac{\pi }{2}$

Minimum is -1 at $-\frac{\pi }{2}\text{ and }\frac{3\pi }{2}$

And zeros at $0,\pm \pi ,\pm 2\pi$

Now use the sense transformation for the graph of $y=\frac{1}{2}\sin x$

:

Maximum is $\frac{1}{2}$ at $-\frac{3\pi }{2}\text{ and }\frac{\pi }{2}$

Minimum is $-\frac{1}{2}$ at $-\frac{\pi }{2}\text{ and }\frac{3\pi }{2}$

And zeros at $0,\pm \pi ,\pm 2\pi$