$ f(x) = -\ln\left(\tanh\frac{x}{2}\right) = \ln \frac{e^{x}+1}{e^{x}-1} $ prove $f(x) = f^{-1}(x)$

Question

$$ f(x) = -\ln\left(\tanh\frac{x}{2}\right) = \ln \frac{e^{x}+1}{e^{x}-1} $$ Prove $f(x) = f^{-1}(x)$ when $x\gt 0$.

I tried to do $f^{-1}(x) = \dfrac 1 {f(x)}$.

Can you help me ?

Answer 1

The inverse function, if it exists, of a mapping $f: S \to T$ is the function $f^{-1}: T \to S$ with the following property.

$$f \circ f^{-1} (x) = x \text{ for all $x$ in $T$}$$

$$f^{-1} \circ f (x) = x \text{ for all $x$ in $S$}$$

To check if functions are inverses, compose $f$ with $f^{-1}$, and vice versa, and see if you get identity. Though here, you of course only need to compose $f$ with itself once.

In general, $f^{-1}(x) \ne \dfrac 1 {f(x)}$. The notation is not to be interpretated as a multiplicative inverse, but rather an inverse mapping.

Another way is to set $f(x) = y$ and solve for $x = f^{-1}(y)$. For this problem, you would see if $f^{-1}(y) = f(x)$. The disadvantage of this second method is that it assumes the inverse exists, whereas the first method will fail if the inverse does not exist.