Question

Find the domain of each of the following functions:

$$\begin{gathered} \left(\mathrm{i}\right)f\left(x\right)={\sin }^{-1}{x}^2 \\ \left(\mathrm{ii}\right)f\left(x\right)={\sin }^{-1}x+\sin x \\ \left(\mathrm{iii}\right)f\left(x\right){\sin }^{-1}\sqrt{x^2-1} \\ \left(\mathrm{iv}\right)f\left(x\right)={\sin }^{-1}x+{\sin }^{-1}2x \end{gathered}$$

Answer 1

(i)
To the domain of ${\sin }^{-1}y$ which is [−1, 1]
$\therefore x^2\in \left[0,1\right]$ as x² can not be negative
$\therefore x\in \left[-1,1\right]$
Hence, the domain is [−1, 1]

(ii)
Let f(x) = g(x) + h(x), where g(x)=cotx and h(x)=cot−1x
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 (−∞, ∞)
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]
$\therefore x^2-1\in \left[0,1\right]$ as square root can not be negative
$$\begin{gathered} \Rightarrow x^2\in \left[1,2\right] \\ \Rightarrow x\in \left[-\sqrt{2},-1\right]\cup \left[1,\sqrt{2}\right] \end{gathered}$$
Hence, the domain is $\left[-\sqrt{2},-1\right]\cup \left[1,\sqrt{2}\right]$

(iv)
Let f(x) = g(x) + h(x), where g(x)=cotx and h(x)=cot−1x
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 $\left[-\frac{1}{2},\frac{1}{2}\right]$