Answer
See explanation for complete proof.
Work Step by Step
Suppose we are given f(x) = an injective(on-to-one) function in x .
Let $y=f(x)$. Swap x and y in this equation. Now we have $x=f(y)$. Solve this for y to get $y=f^{-1}(x)$.
Note that this new relation will be a function because our original function was one-to-one and what was the range of $f$ is now the domain of $f^{-1}$.
We can also convince ourself that
$f(f^{-1}(x)) = f^{-1}(f(x))= x$
Example:
Let $f(x)=2x+3.$
Put $y= f(x)$
$y = 2x+3$
Swap $x$ and $y$
$x=2y+3$
Solve for y
$y=\frac{x-3}{2}$
Thus, $f^{-1}(x) = \frac{x-3}{2}$
