I think that this question was answered in the last couple of days already. (By someone else)
BUT
![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 $f^{-1}(0)+f^{-1}(6)$.](https://data.artofproblemsolving.com/aops20/latex/images/5488810e07ddad30fa1f47cccdcdb9ce71f6a01f.png)
y=3-x This is a line with gradiene = -1 so when it is reflected across y=x it stays the same
So its inverse is y=3-x
$$f(0)=3\qquad so \qquad f^{-1}(0)=3$$
$$\\y=-x^3+2x^2+3x\\
inverse\;\; function\\
x=-y^3+2y^2+3y\\
when\;\; x=6\\
6=-y^3+2y^2+3y\\
0=-y^3+2y^2+3y-6\\
0=-y^2(y-2)+3(y-2)\\
0=(-y^2+3)(y-2)\\
3-y^2=0\qquad or \qquad y-2=0\\
y^2=3\qquad \qquad or \qquad y=2\\
y=\pm\sqrt{3}\qquad \qquad or \qquad y=2\\
y=\pm1.732 \qquad \qquad or \qquad y=2\\
but\;\; y>3 \\
$So no solutions to $ f^{-1}(6)$$
NO SOLUTION.
I have replace the graph the original one had the wrong coefficients and caused me great confusion.
https://www.desmos.com/calculator/alqdacy7uh