Author 
Topic 

mathphobic
Average Member
USA
8 Posts 
Posted  09/08/2007 : 17:37:11

I have this problem but I'm not sure how to do it.
given that f(x) = e^x  e^(x), find the inverse if it exists.
your supposed to see if the function is onetoone. so I put two values in (a and b) and set them equal.
e^a  e^(a) = e^b  e^(b)
I keep ending up with e^b * (e^(2a)  1) = e^a * (e^(2b)  1) and I don't know what to do then.
if anyone could point me in the right direction I'd appreciate it.
thanks.



sahsjing
Advanced Member
USA
2399 Posts 
Posted  09/08/2007 : 18:45:59

Ideas: Let u = e^x. y = u  1/u



mathphobic
Average Member
USA
8 Posts 
Posted  09/08/2007 : 22:13:10

here is what I got
u  1/u = u  1/u
u^2  1 = u^2  1
u^2 = u^2
e^(2x) = e^(2x)
then I put in a and b and it comes out a = b, so it is onetoone.
so then I have to find its inverse.
y = e^x  1/e^x
I don't really know what to do again



mathphobic
Average Member
USA
8 Posts 
Posted  09/09/2007 : 16:34:46

thank you for the help and for being so patient i've always had a hard time with math.
ok e^(2x)  ye^x  1 = 0
solving for e^x
e^x = (y + sqrt(y^2 + 4)/2
taking log of both sides
x = ln ([y + sqrt(y^2 + 4)]/2)
which means for a given output y, there are two inputs x? so it's not onetoone? although looking at the graph it looks like it would be.
y = ln ([x + sqrt(x^2 + 4)]/2)
if you look at the x + sqrt version it looks on the graph like its inverse. and the other version (x  sqrt) is never real ( as far as I can tell). so is the x  sqrt just an extraneous solution or something?
i'm really not trying to be a pest here, i'm just trying to make sure i understand.



peppers18
Advanced Member
Canada
124 Posts 
Posted  09/09/2007 : 17:03:16

it may look like there are 2 solutions but look carefully. check out the one with the 
for all x, x^2+4 is bigger than x^2 so sqrt(x^2+4) is bigger than x so xsqrt(x^2+4) is negative, so you can't take the log of it. so that one gives no solutions. try graphing just that one! it will not graph 


mathphobic
Average Member
USA
8 Posts 
Posted  09/09/2007 : 18:51:18

just out of curiosity, is this a case where the domain of the function is being restricted to just the reals in order for it to be onetoone? if you allowed all the complex numbers the result would mean it wasn't one to one, right? 


sahsjing
Advanced Member
USA
2399 Posts 
Posted  09/09/2007 : 18:55:04

quote: Originally posted by mathphobic
here is what I got
u  1/u = u  1/u
u^2  1 = u^2  1
u^2 = u^2
e^(2x) = e^(2x)
then I put in a and b and it comes out a = b, so it is onetoone.
so then I have to find its inverse.
y = e^x  1/e^x
I don't really know what to do again
y = u  1/u, u > 0 u  yu  1 = 0 u = e^{x} = [y + (y + 4)]/2 x = ln{[y + (y + 4)]/2} Switch x and y, y = ln{[x + (x + 4)]/2} 
Edited by  sahsjing on 09/09/2007 19:06:26 



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