A special function

is not correct. For x, you get . You're supposed to get x.

Actually, it was f(x)=1 and f⁻¹(x) is the line x=1, so f⁻¹(x)≠1.

I did note that x=±1 are not true inverses of f(x)=±1, because x=±1 are vertical lines. However, here I used the fact that an inverse is really just a reflection of f(x) over the line f(x)=x, because it would be "silly" to exclude everything that is not one to one but can be represented in some way (such as parametrically).

x=1 is a reflection of f(x)=1 over the line f(x)=x, and even though f(f⁻¹(x))=1, keep in mind that the inverse of f(x) is the line x=1, which means that f(f⁻¹(x))=1=x. Better yet, we can use the limit process to see that it actually equals x. Let f(x) = Cx+h, where C is some constant. The inverse of f(x) is therefore f⁻¹(x) = (x-h)/C. Now observe that:

f(f⁻¹(x)) = f⁻¹(f(x)) = x

If we take the limit as C approaches 0, f(x) becomes the line f(x)=h and f⁻¹(x) becomes the line x=h. Therefore, technically this works. It's just not immediately apparent.

While this doesn't seem to satisfy the equation, we can observe that even if f⁻¹(x) is x=1 cannot truly be represented, it has a "form" of f⁻¹(x)=∞-y, where y is some impossible y-intercept. To see this, we can consider that even "better" technical solutions are:

1. f(x)=0 and f⁻¹(x)=∞, or x=0
2. f(x)=∞, or x=0, and f⁻¹(x)=0

Since it isn't too difficult to see that 1/∞=0, and vice versa.