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Topic review (newest first)
- mikau
- 2005-10-18 08:41:17
Aww.... lol.
- lifflander
- 2005-10-17 18:24:03
Thank you very much.
- Flowers4Carlos
- 2005-10-17 16:35:34
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!!!
- mikau
- 2005-10-17 14:18:34
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.
- lifflander
- 2005-10-17 14:13:57
also I don't know if the restriction has something to do with it: x >= -1/2
- lifflander
- 2005-10-17 14:12:52
Thanks
- mikau
- 2005-10-17 14:10:05
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.
- lifflander
- 2005-10-17 13:39:41
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.
- mikau
- 2005-10-17 13:34:54
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.
- lifflander
- 2005-10-17 13:29:39
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.