What's the inverse of g(x) = 3 + x + e^x

**TGS** · September 20th 2009, 01:32 PM

g(x) = 3 + x + e^x

Help please?

**skeeter** · September 20th 2009, 01:46 PM

Originally Posted by TGS

g(x) = 3 + x + e^x

Help please?

is this the original question, or did it arise from working out something else?

**TGS** · September 20th 2009, 01:48 PM

its the original question, but i just solved it so nevermind.

Thanks though.

**skeeter** · September 20th 2009, 01:51 PM

Originally Posted by TGS

its the original question, but i just solved it so nevermind.

Thanks though.

is that so ? I would love to see how you found $g^{-1}(x)$ .

**bkarpuz** · September 20th 2009, 01:52 PM

Originally Posted by skeeter

is that so ? I would love to see how you found $g^{-1}(x)$ .

me too

**shawsend** · September 20th 2009, 02:05 PM

$y=3+x+e^{x}$

$e^{y-3}=e^{x+e^x}$

Now putting it in standard Lambert-W terms:

$e^{y-3}=e^x e^{e^x}$

so:

$e^x=W(e^{y-3})$

but:

$y-3-x=e^{x}$

then:

$x=y-3-W(e^{y-3})$

. . . end special function discrimination. Equal rights for special functions.

**skeeter** · September 20th 2009, 02:14 PM

I was hoping for a non-Lambert W function algebraic miracle.

**Atinus** · September 22nd 2009, 04:54 AM

Can someone please provide a simple method to determine this answer? I don't recall how to deal with two variables in an inverse situation especially with one as an exponent. Thanks!

**mr fantastic** · September 22nd 2009, 02:31 PM

Originally Posted by Atinus

Can someone please provide a simple method to determine this answer? I don't recall how to deal with two variables in an inverse situation especially with one as an exponent. Thanks!

If you read this thread while awake you would realise that a 'simple method' is not possible (unless you consider using the Lambert W-function simple).

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