Question:
Find the domain and the range of each of the following real
function: $f(x)=\frac{1}{2-\sin 3 x}$
Solution:
Given: $f(x)=\frac{1}{2-\sin 3 x}$
Need to find: Where the functions are defined.
The maximum value of an angle is $2 \pi$
So, the maximum value of $x=2 \pi / 3$.
Whereas, the minimum value of $x$ is 0
Therefore, the domain of the function, $\mathrm{D} \mathrm{f}(\mathrm{x})=(0,2 \pi / 3)$.
Now, the minimum value of $\sin \theta=0$ and the maximum value of $\sin \theta=1 .$ So, the minimum value of the denominator is 1 , and the maximum value of the denominator is $2 .$
Therefore, the range of the function, $R_{f(x)}=(1 / 2,1)$.