Answer
$f^{-1}_1(x)=-\sqrt{4-x}$ (Red dotted line graph in the image), which is inverse of the function $f(x)$ with domain $(-\infty \lt x \leq 0)$ (Red line graph in the image)
$f^{-1}_2(x)=\sqrt{4-x}$ (Blue dotted line graph in the image), which is inverse of the function $f(x)$ with domain $(0 \leq x \lt \infty)$ (Blue line graph in the image)
Work Step by Step
We know that due to the definition of the inverse function, for a function to have inverse it should have the form that every $y$ value should get only one $x$ value, so we have to restrict either left-hand side(Red on the graph) or right-hand side(Blue on the graph).
So the domain of definition should be:
$(-\infty \lt x \leq 0)$ - Which is Red line graph in the image. Let's call this $f_1(x)$
or
$(0 \leq x \lt \infty)$ - Which is Blue lien graph in the image. Let's call this $f_2(x)$
To calculate inverse of the function $f(x)=4-x^2$, we will follow the steps which is required to calculate it:
First write the function in terms of $y$ and $x$:
$y=4-x^2$
Then replace $y$ by $x$ and vice versa:
$x=4-y^2$
And at last, solve it for $y$:
$y^2=4-x$
$y_1=-\sqrt{4-x}$; $y_2=\sqrt{4-x}$
So we have:
$f^{-1}_1(x)=-\sqrt{4-x}$, which is inverse of the function $f(x)$ with domain $(-\infty \lt x \leq 0)$
(Red dotted line graph in the image)
$f^{-1}_2(x)=\sqrt{4-x}$, which is inverse of the function $f(x)$ with domain $(0 \leq x \lt \infty)$
(Blue dotted line graph in the image)
