MHB What are the domain and range for the inverse function $f^{-1}$ of $f$?

[karush]

Find the inverse function $f^{-1}$ of $f$.
Find the range of $f$ and the domain and range of $f^{-1}$
$$f(x)=5 \sin x +2; \quad -\frac{\pi}{2}\le x \le \frac{\pi}{2}$$
exchange x for y and solve for y
\begin{align*}\displaystyle
x&=5\sin y + 2 \\
\frac{x-2}{5}&=\sin y
\end{align*}
multiply $\sin^{-1}$ to both sides
$$\sin^{-1}\left( \frac{x-2}{5}\right)
=\sin^{-1}(\sin y)=f^{-1}(x)$$

ok how do get the range of $f$ and domain and range of $f^{-1}$

 

[joypav]

Recall that the domain of $f(x)$ is the range of $f^{-1}(x)$, and the range of $f(x)$ is the domain of $f^{-1}(x)$.

So, you are given that the domain of $f(x)$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$.
Therefore, the range of $f^{-1}(x)$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$

Use this idea to determine the range of $f(x)$ and domain of $f^{-1}(x)$. (they should be the same thing)

We know that,
$-1 \leq sinx \leq 1$
If that is the range for $sinx$ then how will that change for $5sinx+2$?

 

[HOI]

A technical point: please don't say "multiply sin^{-1} to both sides". That sounds like you think "sin^{-1}" is a number. What you mean is "apply the function sin^{-1} to both sides".