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 Algebra
 inverse of function
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mathphobic
Average Member

USA
8 Posts

Posted - 09/08/2007 :  17:37:11  Show Profile
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  Show Profile
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  Show Profile
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  Show Profile
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  Show Profile
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  Show Profile
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  Show Profile
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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