Question

Find the domain of the following functions: \[f\left( x \right)={{\sin }^{-1}}{{x}^{2}}\].

Answer 1

Hint:The inverse function of \[\sin x\] is defined for \[-1\le x\le 1\], we need to find if \[{{\sin }^{-1}}{{x}^{2}}\] obeys the same interval. Solve the inequality to find the domain by combining it with the original interval.

Complete step-by-step answer:
We know the inverse sine function \[\left( {{\sin }^{-1}}x \right)\]is defined as \[-1\le x\le 1\](sine only takes on those values). So we know that our function is only defined when \[{{x}^{2}}\]obeys the same interval. We can figure out by solving the following inequality.
 \[-1\le {{x}^{2}}\le 1\]
\[-1\le {{x}^{2}}\], is true for all real numbers, \[x\in R\].
\[{{x}^{2}}\le 1\], can be solved by taking the square root on both sides.
\[\Rightarrow \sqrt{{{x}^{2}}}\le \sqrt{1}\Rightarrow x\le 1\]
Now, by combining with the original interval, we can make the most restrictive of the bonds:
\[\left\{ x|-1\le x\le 1 \right\}\]
\[\therefore \]The function is defined and we get our domain as
\[\left\{ x|-1\le x\le 1 \right\}\]
And \[{{\sin }^{-1}}\]functions always ranges from \[\dfrac{-\pi }{2}\]to\[\dfrac{\pi }{2}\].
[Negative values are only produced by negative input to the function, the \[{{x}^{2}}\]make it so that we will only get positive values on zero.]
\[\therefore \]Range\[=\left\{ y|0\le y\le \dfrac{\pi }{2} \right\}\]

Note: The domain of a function is the set of all possible inputs for the function.The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain.Combine the original intervals and the solved inequalities to find the domain ,we get our domain as \[\left\{ x|-1\le x\le 1 \right\}\] and the range is \[\left\{ y|0\le y\le \dfrac{\pi }{2} \right\}\].