Let $f$ be defined by \[f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\]Calculate $f^{-1}(0)+f^{-1}(6)$.
My go at it:
First I attacked the $f^{-1}(0)$, I found the inverse which turned out to be exactly the same as the non inverse function so because 0 is less than 3, 3-0=3,
so it is $0+f^{-1}(6)$
$f^{-1}(6)$ is the $-x^3+2x^2+3x$, maybe it's a coincident but it factors into $-x(x+1)(x-3)$
