Inverse functions

[Dr Finlay]

Would the inverse function of

 

[math] \displaystyle\frac{1 + e^x}{1 - e^x}[/math]

 

be

 

[math]\displaystyle\frac{e^x - 1}{e^x + 1}[/math]

?

 

Just been working through Stewarts calculus, the book only gives answers to the odd numbered questions. Most of the concepts in the book are pretty new to me, having only just started my Alevels, so i thought it would be best to get as much practice at it as i could.

 

Cheers,

Rob

[Ducky Havok]

No, that is not the inverse. I believe all you did was say 1 over the function, right? The way to do inverses is to switch your x's and y's in the equation, and the solve for y.

 

[math]x=\frac{1+e^y}{1-e^y} [/math]

[math]x-xe^y=1+e^y[/math]

[math]x-1=(x+1)e^y[/math]

[math]\frac{x-1}{x+1}=e^y[/math]

[math]\ln{\frac{x-1}{x+2}}=y[/math]

or

[math]\ln{(x-1)}-\ln{(x+2)}=y [/math]

[Dr Finlay]

Cheers mate