Finding the Inverse of a Function: Solving for y=x+√(x^2-1)

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Homework Statement



I need to find the inverse of y=x+√(x^2-1)

Homework Equations





The Attempt at a Solution



I know it's undefined from x=-1 and x=1 so there must be two different inverse functions on two different intervals. I don't know how to find them though.
 
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You find the inverse pretty much the way you find any inverse:
Given y= x+ \sqrt{x^2- 1}, solve for x. Since there is a square root, we will want to square, and, in order not to get another square root in the "cross term" we want it by itself: y- x= \sqrt{x^2- 1}. Now square that and solve for x.

Note the while x cannot be between -1 and 1, y can go to 0.
 
anthonych414 said:

Homework Statement



I need to find the inverse of y=x+√(x^2-1)

Homework Equations



The Attempt at a Solution



I know it's undefined from x=-1 and x=1 so there must be two different inverse functions on two different intervals. I don't know how to find them though.
Let's call the function you are given, f, so that
f(x)=x+\sqrt{x^2-1}\ .​

Following HallsofIvy's suggestion you will find:
x=g(y)\ .​
For the function, g, to be the inverse of function, f, it must also be true that g is the inverse of f. However, the domain of g will need to be restricted (to the range of f) so that g is a 1 to 1 function.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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