Suppose that we have function $y=\sin(x)$ we need to find its inverse function, assuming that $D(f)=[-\pi/4.\pi/4]$
I know that inverse of $\sin(x)$ is $\arcsin(x)$, it would be answer of a given function too, but why do I need $D(f)=[-\pi/4.\pi/4]$? I don't know, should I introduce some variable $c$, so that $y=\sin(x)$ will look like $y=\sin(x-c)$ or $y=\sin(x)+c$? Please give me a hint. In case I meet similar problem, like "find inverse of function $y=f(x)$ where $D(f)=[a,b]$" what should I do? As I know domain of given function and range of inverse function are the same, so it means that the range of $\arcsin(x)$ is $[-\pi/4.\pi/4]$, but how to proceed?
Best Answer
The typical range of $\arcsin(x)$ is $[-\pi/2,\pi/2]$, with domain $[-1,1]$. This is because we restrict $\sin(x)$ to the interval $[-\pi/2,\pi/2]$--a maximal interval on which $\sin(x)$ is one-to-one--and then take the inverse.
However, we didn't have to have a maximal interval, just some interval on which it's one-to-one, so the same principle will apply here, and you'll simply have $f^{-1}(x)$ as the restriction of $\arcsin(x)$ to the interval $[-1/\sqrt{2},1/\sqrt{2}]$.