Find a formula for the inverse of this function

[jwpaine]

Find a formula for the inverse of y = ( 1 + e^x ) / ( 1 - e^x )
Here is my work and then I have a question at the end.

So I cross multiplied:

y( 1 - e^x ) = (1 + e^x)

distributed the y on the left side:

y - e^x = 1 + e^x

added e^x to both sides:

y = 1 + 2(e^x)

re-arranged to get e^x on one side:

y - 1 = 2(e^x)

divided both sides by 2

(y - 1 )/2 = e^x

and then lned both sides to get x without the natural base e:

ln((y - 1 )/2) = x


is this correct?
My Calculus professor says:

[quote:2hd63i1a]So I cross multiplied:
y( 1 - e^x ) = (1 + e^x)
distrubuted the y on the left side:
y - e^x = 1 + e^x

You need to distribute y into all terms of which it is a factor.

[/quote:2hd63i1a]

 

[pka]

\(\displaystyle \L y\left( {1 - e^x } \right) = y - ye^x\)

 

[stapel]

y( 1 - e^x ) = (1 + e^x)

distributed the y on the left side:

y - e^x = 1 + e^x

You might want to check your work on the left-hand side:

. . . . .y(1 - e^x) = 1 + e^x

. . . . .y(1) + y(-e^x) = 1 + e^x

. . . . .y - e^xy = 1 + e^x

...and so forth.

I think you lost a "y" somewhere...?

Eliz.

 

[jwpaine]

Of course.... how silly of me.. thanks!