Inverse function of $x^2-4$ The function $h$ is defined by $$h(x)=x^2-4$$. for $$x\leq0$$
Find an expression for $h^{-1}(x)$
My attempt, 
Let $h^{-1}(x)=a$
$x=h(a)$
$x=a^2-4$
$a=\sqrt{x+4}$
$h^{-1}(x)=\sqrt{x+4}$
Am I wrong? Is the answer $-\sqrt{x+4}$?

Other answers have already correctly addressed it, but just for the sake of having another point of view, here's how I'd run through an example when I used to teach it.
We're given:
$$ h(x) = x^2 - 4, \qquad x \le 0 $$
Replace $h(x)$ with $y$:
$$ y = x^2 - 4, \qquad x \le 0 $$
Swap all $x$ and $y$, including in the restriction:
$$ x = y^2 - 4, \qquad y \le 0 $$
Notice now we have $y \le 0$ since we swapped $x$ and $y$ in the restriction!
The next step is to solve for $y$ in the equation:
\begin{align}
  x &= y^2 - 4\\
  x+4 &= y^2\\
  \sqrt{x+4} &= \sqrt{y^2}\\
  \sqrt{x+4} &= |y|\\
  \pm \sqrt{x+4} &= y
\end{align}
As others have pointed out, we must be careful and remember that $\sqrt{y^2} = |y|$ in general.  The very last equality above, i.e., $y = \pm\sqrt{x+4}$, follows from the definition of absolute value.
So now the question is, which root do we take?  The positive or negative root?  Well, remember our restriction earlier.  It says $y \le 0$.  Therefore we must take the negative root:  $y = -\sqrt{x+4}$.  So we have:
$$ y = -\sqrt{x+4}, \qquad y \le 0 $$
Finally, replace $y$ with $h^{-1}(x)$ to get the final result:
$$h^{-1}(x) = -\sqrt{x+4}, \qquad h^{-1}(x) \le 0$$
Notice how the domain of $h(x)$ and range of $h^{-1}(x)$ are the same, as we would expect.  They're both  $(-\infty, 0]$ in interval notation.
