Answer
$$
f(x)=\sqrt{x+3}, \quad x \geq -3,
$$
we find that:
$$
f^{-1}(x)=x^{2}-3 , \quad x \geq 0.
$$
Work Step by Step
we first write
$$
y=f(x)=\sqrt{x+3}, \quad x \geq -3.
$$
Then we solve this equation for $ x$ as a function of $y$
$$
y^{2}= x+3
$$
$$
x =y^{2}-3
$$
which tells us that
$$
f^{-1}(y)=y^{2}-3. \quad \quad (i)
$$
Since we want $x$ to be the independent variable, we reverse $x$ and $y $ in (i) to produce the formula
$$
f^{-1}(x)=x^{2}-3.
$$
We know the domain of $f^{ −1}$ is the range of $f$
whereas the range of $f(x)=\sqrt{x+3} $ is $ [0, \infty) $. Thus, if we
want to make the domain of $f^{ −1}$ clear, we must express it explicitly by rewriting (i) as
$$
f^{-1}(x)=x^{2}-3 , \quad x \geq 0.
$$
