Answer
$f^{-1} (x)=\dfrac{x^2-3}{2}$
The domain for the function is $x \geq 0$
Work Step by Step
When we apply the horizontal test, it has been noticed that the function is one-to-one and verifies the horizontal test.
Therefore, the function has an inverse function.
To compute the inverse, we will have to interchange $x$ and $y$.
$x=\sqrt {2y+3} \implies x^2=2y+3$
or, $y=\dfrac{x^2-3}{2}$
Replace $y$ with $f^{-1} (x)$.
so, $f^{-1} (x)=\dfrac{x^2-3}{2}$
The domain for the function is $x \geq 0$.
