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