Inverse Function problem involving e^x

  • #1

Homework Statement



Let g(x) = (e^x - e^-x)/2. Find g^-1(x) and show (by manual computation) that g(g^-1(x)) = x.

Homework Equations


g(x) = (e^x - e^-x)/2

The Attempt at a Solution



I get the inverse = ln[ (2x + sqrt(4x^2 + 4) ) / 2 ]

How do I proceed?
 
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  • #2
First of all,

[tex]\frac{2x\pm \sqrt{4x^2+4}}{2}[/tex] can be simplified. Factor out 4 from the square root and cancel the 2's.

If g(x) = x then what is g(2x)? Then what is g(f(x))?
 
  • #3
Thank you very much. I understood the question and answered it

PS: Sorry for the late reply.
 
  • #4
ChromoZoneX said:

Homework Statement



Let g(x) = (e^x - e^-x)/2. Find g^-1(x) and show (by manual computation) that g(g^-1(x)) = x.

Homework Equations


g(x) = (e^x - e^-x)/2

The Attempt at a Solution



I get the inverse = ln[ (2x + sqrt(4x^2 + 4) ) / 2 ]

How do I proceed?

First: please use brackets, so write e^(-x) instead of e^-x and g^(-1) instead of g^-1 (however, e^x is OK as written). You want to find what x gives you g(x) = y; that would be g^(-1)(y). Just put z = e^x, so you have the equation (1/2)(z + 1/z) = y, which is solvable for z. After that, x = log(z).

BTW: to guard against confusing yourself and others, I suggest you refer to the argument of the inverse function as y (or z, or w or anything different from x), at least until you have obtained the final result. Then you can switch to any symbol you want. However, if your teacher wants you to do it another way, follow the requirements you are given.

RGV
 

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