How do you find the inverse of #f(x)=x^2 - 1# and is it a function?

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Answer 1 · 2016-03-23T22:55:43

Let #y = f(x)# and solve for #x#.

We find that there is no inverse function unless the domain of #f(x)# is restricted.

Suppose #f(x) = x^2-1#

To attempt to find an inverse function, let #y = f(x)# and solve for #x# in terms of #y#...

#y = x^2-1#

Add #1# to both sides to get:

#y + 1 = x^2#

Transpose and take square root of both sides, allowing for either sign:

#x = +-sqrt(y+1)#

This does not determine a unique value for #x# in terms of #y#. So there is no inverse function, unless we restrict the domain of #f(x)#.

For example, if we specify an explicit domain #[0, oo)# for #f(x)#, then #f^(-1)(y) = sqrt(y+1)#

Alternatively, we might specify an explicit domain #(-oo, 0]# for #f(x)#, then #f^(-1)(y) = -sqrt(y+1)#