+17 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 $f^{-1}(0)+f^{-1}(6)$.
+128732 f-1(0)
Set the first function = 0
0 = 3- x
-3 = -x
x = 3
So f(3) = 0
Which means that f-1(0) = 3
f-1(6)
Set the second function = 6
-x^3 + 2x^2 + 3x = 6
By inspection , x = 2 ..... no good....x must be > 3
So set the first function = 6
6 = 3 - x
x = 3 - 6
x = -3
So f(-3) = 6
Which means that f-1(6) = -3
So
f-1(0) + f-1(6) = 3 + - 3 = 0
