#9**+5 **

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?!!!!!

.

Alan
Jan 11, 2015

#1**+5 **

I think that this question was answered in the last couple of days already. (By someone else)

BUT

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.

Melody
Jan 10, 2015

#2**+5 **

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....

CPhill
Jan 10, 2015

#3**+5 **

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?**

------------------------------

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.

----------------------------------

Melody
Jan 10, 2015

#4**+5 **

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 ?????

CPhill
Jan 10, 2015

#5**+5 **

Yes i am happy with my answer I think.

I have played with the graph some more as well and I am totally happy.

Melody
Jan 10, 2015

#6**+5 **

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...!!!]

CPhill
Jan 10, 2015

#7**0 **

I think 6 * is* in the domain of the inverse function. When f(x) = 6, x = -3, so f

.

Alan
Jan 10, 2015

#8**+5 **

Okay Alan can you please spot the fault in my logic?

$$\\y=-x^3+2x^2+3x \qquad for \;\;x>3\\\\

$Inverse function$\\

x=-y^3+2y^2+3y \qquad for \;\;y>3\\\\

When\;\;x=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.

Melody
Jan 11, 2015

#9**+5 **

Best Answer

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?!!!!!

.

Alan
Jan 11, 2015