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Posted by Soroban on September 11, 2002 at 13:50:38:

: Hello, I hope someone can help me with this.

: f(x)=(1+3x)/(2-2x)

First, I'd let y = f(x). It makes for neater algebra.

: I have to find the formula for the inverse function.
: I get x= (-2*f(x)-1)/(s*f(x)-3), but it is wrong.

This would be: x = (-2y-1)/(2y-3)
I assume that "s" is a "2".

We have: y = (1 + 3x)/(2 - 2x)

: I divided both sides by (2-2x)
You meant "multiplied", of course.
y(2 - 2x) = 1 + 3x

: I distributed
2y - 2xy = 1 + 3x

: I isolated for x terms
-2xy - 3x = 1 - 2y

: I factored out x
x(-2y - 3) = 1 - 2y

: I divided both sides by coefficient of x
x = (1 - 2y)/(-2y - 3)

Multiply numerator and denominator by -1:
x = (2y - 1)/(2y + 3)

Finally: f-inverse(x) = (2x - 1)/(2x + 3)

: Any ideas what I'm doing wrong?

You seem to have dropped (on inserted) a couple of minus-signs.
It may have happened when you "isolated for x terms".

: Thanks, Larry

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