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**student007****Member**- Registered: 2005-12-13
- Posts: 2

Find inverse of this function

F(X) = X/3X-2

Interchange X & Y

Y = X/3X-2

X = Y/3Y-2

X(3Y-2) = Y

3XY-2X = Y

I'm not sure what to do from here any help would be appreciated. Thanks

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

x(3y - 2) = y

x = y / (3y - 2)

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**student007****Member**- Registered: 2005-12-13
- Posts: 2

thank you

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Whoops, you used a different way of finding the inverse than I'm used to, so I got mixed up.

What I normally do is:

y = x/(3x - 2)

y(3x - 2) = x

3xy - 2y = x

3xy - x = 2y

x(3y - 1) = 2y

x = 2y / (3y - 1)

What you did was pretty much the same, but you switched the x and y variables, as it is standard to want to solve for y:

x = y/(3y - 2)

x(3y - 2) = y

3xy - 2x = y

3xy - y = 2x

y(3x - 1) = 2x

y = 2x / (3x-1)

As you can see, you come up with the same answer, just different names for the variables. In mine, x is dependant on y, in yours (as it normally is), y is dependant on x.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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