Chapter 4.6, Problem 67E

Solution

To determine : The domain and range of the function $f\left(x\right)=\sqrt{\sin x}$ .

The domain of the given function is $\left(2n\pi ,\left(2n+1\right)\pi \right)$ and its range is $\left[0,1\right]$ .

Given information :

The function $f\left(x\right)=\sqrt{\sin x}$ .

Concept used:

The domain of a function is the set of all the real values for which the function is meaningful and range of a function is the set of all real values taken by the function at points in its domain.

Explanation :

Consider the given function $f\left(x\right)=\sqrt{\sin x}$ .

Graph the given function in the viewing window $\left[-4\pi ,4\pi \right]$ by $\left[-7,7\right]$ .

  

From the graph it is evident that the function is defined for all values of $x$ , which lie between an integral multiple of $\pi$ and the next integral multiple but it is not defined for the next interval and it continues alternately in that way. The function assumes all real values between $0\text{ and }1$ .

Therefore, the domain of the given function is $\left(2n\pi ,\left(2n+1\right)\pi \right)$ and its range is $\left[0,1\right]$ .