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Guest Jan 9, 2015

#9
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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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Alan  Jan 11, 2015
#1
+94105
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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.

https://www.desmos.com/calculator/alqdacy7uh

Melody  Jan 10, 2015
#2
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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
+94105
+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 !

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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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Melody  Jan 10, 2015
#4
+92650
+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
+94105
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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
+92650
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[ 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
+27227
0

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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Alan  Jan 10, 2015
#8
+94105
+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

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
+27227
+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
#10
+94105
0

Yes it is totally confusing!     Thanks Alan,  I will try and work out what you are saying.

Melody  Jan 11, 2015