Answer to Question #250135 in Calculus for Alunsina

Question

Find the inverse of the given function.

h(x)=sin^-1 sqrt x

Answer 1

h(x)= $sin^{-1}(\sqrt{x})$

Inverse of h(x) will exist if it is bijective function.

So Domain of h(x) is [0,1] and range is [0,$\frac{π}{2}]$

Let y = $sin^{-1}(\sqrt{x})$

=> Sin(y) = $\sqrt{x}$

=> x = sin²y

So inverse of h(x) , h$^{-1}(x)=$ sin²(x) with domain [0,$\frac{π}{2}]$ and range [0,1]