Find the inverse for the function `f(x)=sqrt(x^2-1)` :

The given function does not have an inverse as it is not 1-1 (injective.) The graph does not pass the horizontal line test:

(Note that the function is undefined in the reals for `-1<x<1` )

You can create an inverse for the two pieces of the graph:

(1) For x>1 (Since 1 is not in the domain):

`y=sqrt(x^2-1)` Domain is x>1, Range is y>0

`x=sqrt(y^2-1)`

`x^2=y^2-1`

`y^2=x^2+1`

`y=+-sqrt(x^2+1)` Since the range of f(x) is y>0, the domain for the inverse is x>0 and the domain of f(x) is x>1 so the range of the inverse is y>1 so choose `f^(-1)(x)=sqrt(x^2+1)`

(2) For x<-1 we have the domain is x<-1 the range is y>0.

Again we have `y=+-sqrt(x^2+1)` . Here the range of f(x) is y>0 so the domain is x>0 and the domain of f(x) is x<-1 so the range of the inverse is y<-1 so choose `f^(-1)(x)=-sqrt(x^2+1)`

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**The given function does not have an inverse**. If you look at the pieces of the function, for x>1 the inverse would be `f^(-1)(x)=sqrt(x^2+1)` (x>0) and for x<-1 the inverse would be `f^(-1)(x)=-sqrt(x^2+1)` (x>0)

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The graph of the function (black) and the inverses of the pieces in red and blue:

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