+33657 The "cubic" part of the inverse function is defined for y less than 0 not y greater than 3 - see http://web2.0calc.com/questions/math-problem_17#r4
Look at the graph of the original function. It is defined for all x and all y, and has a one-to-one mapping x onto y and a one-to-one mapping y onto x, so the inverse function does also.
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You have interpreted the inverse function by just interchanging x and y. I've interpreted it as as a mapping back from the original f(x) onto the original x, which means, for me, the "x" in f-1(x) is on the original "y" axis, and f-1(x) is on the original "x" axis.
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Is this confusing or what?!!!!!
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+118702 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 $f^{-1}(0)+f^{-1}(6)$.](/api/ssl-img-proxy?src=https%3A%2F%2Fdata.artofproblemsolving.com%2Faops20%2Flatex%2Fimages%2F5488810e07ddad30fa1f47cccdcdb9ce71f6a01f.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)=3sof−1(0)=3
y=−x3+2x2+3xinversefunctionx=−y3+2y2+3ywhenx=66=−y3+2y2+3y0=−y3+2y2+3y−60=−y2(y−2)+3(y−2)0=(−y2+3)(y−2)3−y2=0ory−2=0y2=3ory=2y=±√3ory=2y=±1.732ory=2buty>3$Sonosolutionsto$f−1(6)
NO SOLUTION.
I have replace the graph the original one had the wrong coefficients and caused me great confusion.
+130477 Melody...I don't think it's going to be very easy to "manually" isolate "x" in the function -x^3 -2x^2 + 3x so that an inverse an be found ..!!!!
Here's what WolfamAlpha generated as the inverse.....it looks pretty "nasty".... (to say the least)
I think Alan answered this one.....but he did it graphialy, as I recall....
+118702
I am sticking with my first answer - I think that the algebra is correct
My original graph had the wrong coefficients so it was incorrect ***
This graph is correct and it shows that there are no values for f−1(6)
https://www.desmos.com/calculator/d86qhh4t5m
So I still say there are no solutions. Because x=6 is not in the domain of the invese function !
What answer did Alan get?
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Also Chris the whole graph is of no consequence because there are restrictions on the range of the inverse fuction,
y>3 therefore x<0 therefore f−1(6) does not exist.
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+130477 I did this the same as you and found....as you did...that whatever values were put into the original function to produce "6" - namely, ±√3 and 2 - were outside the prescribed domain of the original function...!!!!
Maybe we have "clunker" here ?????
+118702 Yes i am happy with my answer I think.
I have played with the graph some more as well and I am totally happy.
+130477 Here's Alan's answer......http://web2.0calc.com/questions/math-problem_17
[ I don't see what he did, but I'm pretty sure that you and I are correct....no solution to the second part of the problem as it was presented...!!!]
+33657 I think 6 is in the domain of the inverse function. When f(x) = 6, x = -3, so f-1(6) = -3. This is on the linear part of the function.
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+118702 Okay Alan can you please spot the fault in my logic?
y=−x3+2x2+3xforx>3$Inversefunction$x=−y3+2y2+3yfory>3Whenx=6
I was going to keep going but I have already done it in my first post
http://web2.0calc.com/questions/please-help_65#r1
When x=6 there are 3 solutions for y but for each y is less than 3 so there are no solutions.
NOW
To confirm this I did a graph. The blue graph on the top is the inverse graph.
It confirms that there is no f-1(6)
Which bit of my logic seems wrong to you?
OR is it just that the result disagrees with your logic and you haven't worked out why yet?
I am not trying to put you on the spot but usually you can show me what I have done wrong. 
+33657 The "cubic" part of the inverse function is defined for y less than 0 not y greater than 3 - see http://web2.0calc.com/questions/math-problem_17#r4
Look at the graph of the original function. It is defined for all x and all y, and has a one-to-one mapping x onto y and a one-to-one mapping y onto x, so the inverse function does also.
.
You have interpreted the inverse function by just interchanging x and y. I've interpreted it as as a mapping back from the original f(x) onto the original x, which means, for me, the "x" in f-1(x) is on the original "y" axis, and f-1(x) is on the original "x" axis.
.
Is this confusing or what?!!!!!
.
