Find the domain of each of the following functions:

Solution:

(i)

To the domain of $\sin ^{-1} y$ which is $[-1,1]$

$\therefore x^{2} \in[0,1]$ as $x^{2}$ can not be negative

$\therefore x \in[-1,1]$

Hence, the domain is $[-1,1]$

(ii)

Let $f(x)=g(x)+h(x)$, where

Therefore, the domain of $f(x)$ is given by the intersection of the domain of $g(x)$ and $h(x)$

The domain of $g(x)$ is $[-1,1]$

The domain of $h(x)$ is $(-\infty, \infty)$

Therfore, the intersection of $g(x)$ and $h(x)$ is $[-1,1]$

Hence, the domain is $[-1,1]$.

(iii)

To the domain of $\sin ^{-1} y$ which is $[-1,1]$

$\therefore x^{2}-1 \in[0,1]$ as square root can not be negative

$\Rightarrow x^{2} \in[1,2]$

$\Rightarrow x \in[-\sqrt{2},-1] \cup[1, \sqrt{2}]$

Hence, the domain is $[-\sqrt{2},-1] \cup[1, \sqrt{2}]$

(iv)

Let $f(x)=g(x)+h(x)$, where

Therefore, the domain of $f(x)$ is given by the intersection of the domain of $g(x)$ and $h(x)$

The domain of $g(x)$ is $[-1,1]$

The domain of $h(x)$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$

Therfore, the intersection of $g(x)$ and $h(x)$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$

Hence, the domain is $\left[-\frac{1}{2}, \frac{1}{2}\right]$