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

You are not logged in.

## #1 2005-10-17 13:29:39

lifflander
Novice

Offline

### Help! - Function inverse

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.

Last edited by lifflander (2005-10-17 14:03:58)

## #2 2005-10-17 13:34:54

mikau
Super Member

Offline

### Re: Help! - Function inverse

I think your writing the inverse function, so your actually solving for x and not for y. I'm trying to remember how this is done. I think you swap x and y, and then once you solve for it, mirror it about the line y = x or something like that. I need to review it.

A logarithm is just a misspelled algorithm.

## #3 2005-10-17 13:39:41

lifflander
Novice

Offline

### Re: Help! - Function inverse

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.

## #4 2005-10-17 14:10:05

mikau
Super Member

Offline

### Re: Help! - Function inverse

Because its an exponential equation. I remember this exact topic being discussed but I can't remember how it went. I'll see if I can find it in my mathbook.

A logarithm is just a misspelled algorithm.

lifflander
Novice

Offline

Thanks

## #6 2005-10-17 14:13:57

lifflander
Novice

Offline

### Re: Help! - Function inverse

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

## #7 2005-10-17 14:18:34

mikau
Super Member

Offline

### Re: Help! - Function inverse

Well I have to sign off now, I'm still looking for it. I'll post it tommorow if I find it. Sorry for the delay.

A logarithm is just a misspelled algorithm.

## #8 2005-10-17 16:35:34

Flowers4Carlos
Full Member

Offline

### Re: Help! - Function inverse

hi yaz lifflander!

y = x² + x                       complete the square to obtain
y = (x + 1/2)² - 1/4
x = (y + 1/2)² - 1/4
x + 1/4 = (y + 1/2)²
±(x + 1/4)½ = y + 1/2     since x≥-1/2    y must be positive
(x + 1/4)½ = y + 1/2
(x + 1/4)½ - 1/2 = y

there ya go!!!  take care buddy!!!

## #9 2005-10-17 18:24:03

lifflander
Novice

Offline

### Re: Help! - Function inverse

Thank you very much.

## #10 2005-10-18 08:41:17

mikau
Super Member

Offline

### Re: Help! - Function inverse

Aww.... lol.

A logarithm is just a misspelled algorithm.