Let $f(x) = x|x|.$  Find $f^{-1}(4) + f^{-1}(-100).$
To find the inverse function, I can switch $x$ and $y$ and then solve for $y$: $x = y|y|.$ To get rid of the absolute value sign, I can split the equation into two cases: $y \geq 0$ and $y < 0.$ If $y \geq 0,$ then $|y| = y,$ so I have $x = y^2.$ Taking the square root of both sides, I get $y = \pm \sqrt{x}.$ But since I assumed $y \geq 0,$ I only take the positive root: $y = \sqrt{x}.$ If $y < 0,$ then $|y| = -y,$ so I have $x = -y^2.$ Taking the square root of both sides, I get $y = \pm \sqrt{-x}.$ But since I assumed $y < 0,$ I only take the negative root: $y = -\sqrt{-x}.$ Therefore, the inverse function is a piecewise function: $f^{-1}(x) = \sqrt{x}$ if $x \geq 0$ and $f^{-1}(x) = -\sqrt{-x}$ if $x < 0.$ Now I can evaluate $f^{-1}(4) + f^{-1}(-100).$ Since 4 is non-negative, I use the first formula: $f^{-1}(4) = \sqrt{4} = 2.$ Since -100 is negative, I use the second formula: $f^{-1}(-100) = -\sqrt{-(-100)} = -\sqrt{100} = -10.$ Adding these two values, I get $2 + (-10) = -8.$ 