Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

You are not logged in.

## #1 Re: Help Me ! » Help! - Function inverse » 2005-10-16 20:24:03

Thank you very much.

## #2 Re: Help Me ! » Help! - Function inverse » 2005-10-16 16:13:57

also I don't know if the restriction has something to do with it: x >= -1/2

Thanks

## #4 Re: Help Me ! » Help! - Function inverse » 2005-10-16 15:39:41

But without graphing the equation I should be able to find the inverse - let me give you an example:

f(x) = 2x + 1
y = 2x + 1

switch x and y

x = 2y + 1

x - 1 = 2y

x - 1 / 2 = y

so the f-1(x) = x - 1 / 2

But for some reason I can't figure out how to do that with the problem that I have now.

## #5 Help Me ! » Help! - Function inverse » 2005-10-16 15:29:39

lifflander
Replies: 9

Problem:

f(x) = x^2 + x when x >= -1/2

I know the basic way to find the inverse of a function, but for some reason I can't seem to figure out what I'm doing wrong with this problem. Here's were I was going with it:

f(x) is the same as y.

y = x^2 + x

switch x and y

x = y^2 + y

here's where I get stuck - I'm trying to solve for y, so do I move the y over or something else???

Thanks for you help.