The inverse of a function

Aug 2018
128
7
România
Let $\text{W}$ denote the Lambert W function.

Let $y$ denote the inverse of the function $-x + \log_3(x)$ with domain (0, 1],
then $x = -y + \log_3(y) = -y + \ln(y)/\!\ln(3)$, where $x \leqslant -1$ and $0 < y \leqslant 1$,
so $\ln(3)x = -\ln(3)y + \ln(y)$,
which implies $3^{\large x} = ye^{-\large\ln(3)y}$,
and so $-\ln(3)3^x = -\ln(3)ye^{-\large\ln(3)y}$.
Hence $-\ln(3)y = \text{W}\!\left(-\ln(3)3^{\large x}\right)$,
i.e. $y = -\text{W}\!\left(-\ln(3)3^{\large x})/\!\ln(3\right)$.

If $x \geqslant 1$, the inverse is $-\text{W}_{-1}\!\left(-\ln(3)3^{\large x})/\!\ln(3\right)$.

(I've omitted some justification of the above.)
Hello,

Do I understand, however, that the function has two inverse functions?Thank you very much!

All the best,

Integrator
 

skipjack

Forum Staff
Dec 2006
21,379
2,410
The original function isn't invertible. If its domain is changed (without restricting it more than necessary) to produce an invertible function (there are two ways of doing that), it will have an inverse that's unique, but depends on the domain chosen.
 
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Aug 2018
128
7
România
The original function isn't invertible. If its domain is changed (without restricting it more than necessary) to produce an invertible function (there are two ways of doing that), it will have an inverse that's unique, but depends on the domain chosen.
Hello,

Now , I understand! Thank you very much!

All the best,

Integrator