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mathphobic
Average Member
USA
8 Posts |
 Posted - 09/08/2007 : 17:37:11
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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 one-to-one. 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.
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sahsjing
Advanced Member
USA
2399 Posts |
Posted - 09/08/2007 : 18:45:59
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Ideas:
Let u = e^x.
y = u - 1/u
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mathphobic
Average Member
USA
8 Posts |
Posted - 09/08/2007 : 22:13:10
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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 one-to-one.
so then I have to find its inverse.
y = e^x - 1/e^x
I don't really know what to do again
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mathphobic
Average Member
USA
8 Posts |
Posted - 09/09/2007 : 16:34:46
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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 one-to-one? 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.
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peppers18
Advanced Member
Canada
124 Posts |
Posted - 09/09/2007 : 17:03:16
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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 x-sqrt(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 |
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mathphobic
Average Member
USA
8 Posts |
Posted - 09/09/2007 : 18:51:18
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| 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 one-to-one? if you allowed all the complex numbers the result would mean it wasn't one to one, right? |
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sahsjing
Advanced Member
USA
2399 Posts |
Posted - 09/09/2007 : 18:55:04
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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 one-to-one.
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 = ex = [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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