Chapter 8, Problem 1RE

To determine

The domain and range of $y={\sin }^{-1}x$.

Answer to Problem 1RE

The domain of $y={\sin }^{-1}x$ is the interval $\underset{\_}{\left[-1,1\right]}$ and the range of $y={\sin }^{-1}x$ is $\underset{\_}{\left[-\frac{\pi }{2},\frac{\pi }{2}\right]}$.

Explanation of Solution

Definition used:

Domain:

The domain of a function $y=f\left(x\right)$ is defined as the set of all $x$ values for which the function is defined.

Range:

The range of a function $y=f\left(x\right)$ is defined as the set of all $y$ values that $f$ takes.

Calculation:

Recall the property of functions and their inverses that domain of the function $f$ equals the range of its inverse function $f^{-1}$.

Also the domain of the inverse function $f^{-1}$ equals the range of the function $f$.

Restrict the domain of the function $y=\sin x$ to the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ so that the restricted function $y=\sin x-\frac{\pi }{2}\le x\le \frac{\pi }{2}$ is one to one and has inverse symbolized by $y={\sin }^{-1}x$.

Use the online graphic tool to sketch the graph of the function $y=\sin x-\frac{\pi }{2}\le x\le \frac{\pi }{2}$ as follows:

From the Figure 1 observe that range of $y=\sin x-\frac{\pi }{2}\le x\le \frac{\pi }{2}$, is the interval $\left[-1,1\right]$.

Therefore by the property domain of the inverse function $y={\sin }^{-1}x$ is the interval $\underset{\_}{\left[-1,1\right]}$ and the range is the interval $\underset{\_}{\left[-\frac{\pi }{2},\frac{\pi }{2}\right]}$.