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 $-\sqrt{x+4}$?

There is a misstep in your procedure. 
$y = x^2 - 4 \implies x^2 = y + 4\implies \sqrt{x^2} = \sqrt{y+4}$
Note that $\sqrt x^2 = x$ but $\sqrt{x^2} \neq x$ in general. Actually, $\sqrt{x^2} = |x|$, and combined with $|x| = -x$ for $x\leq 0$ we have
$-x = \sqrt{y+4}\implies x=-\sqrt{y+4}\implies h^{-1}(x) = -\sqrt{x+4}$
Quick check gives us
$h(h^{-1}(x)) = h(-\sqrt{x+4})= (-\sqrt{x+4})^2 - 4 = x + 4 - 4 = x$
$h^{-1}(h(x)) = h^{-1}(x^2 - 4) = -\sqrt{x^2-4 + 4} = -\sqrt{x^2} = - |x| = x\quad (x\leq 0)$
