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## #1 2005-10-16 15:29:39

lifflander
Member
Registered: 2005-10-16
Posts: 5

### 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 (2005-10-16 16:03:58)

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## #2 2005-10-16 15:34:54

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### 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.

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## #3 2005-10-16 15:39:41

lifflander
Member
Registered: 2005-10-16
Posts: 5

### 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.

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## #4 2005-10-16 16:10:05

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### 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.

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## #5 2005-10-16 16:12:52

lifflander
Member
Registered: 2005-10-16
Posts: 5

Thanks

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## #6 2005-10-16 16:13:57

lifflander
Member
Registered: 2005-10-16
Posts: 5

### Re: Help! - Function inverse

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

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## #7 2005-10-16 16:18:34

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### 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.

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## #8 2005-10-16 18:35:34

Flowers4Carlos
Member
Registered: 2005-08-25
Posts: 106

### 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!!!

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## #9 2005-10-16 20:24:03

lifflander
Member
Registered: 2005-10-16
Posts: 5

### Re: Help! - Function inverse

Thank you very much.

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## #10 2005-10-17 10:41:17

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### Re: Help! - Function inverse

Aww.... lol.

A logarithm is just a misspelled algorithm.

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