+865 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 \(f^{-1}(0)+f^{-1}(6)\).
+781 We find the value of f(x) that results in 3(f(x)=0).
For 3-x, x=3. For -x^3+2x^2+3x, x=0, 3, -1. None of these satisfies x>3, so f(0) cannot go into the second piece of the function. It can only go in the first, so f^-1(0) is 3.
Then we find the value of f(x) that results in 6.
For 3-x, x=-3. For -x^3+2x^2+3x, x=2, \(\sqrt{3}\ \text{or}-\sqrt{3}\). None of these satisfies x>3, so again, f(6) cannot go into the second piece of the function. It can only go in the first, so f^-1(6) is -3.
Adding these together, 3+(-3)=0.
