# Prove that the function f(x)=tanh^-1(x) is an odd function ?

## prove that the function f(x)=tanh^-1(x) is an odd function

Apr 19, 2018

The argument below can be adapted to prove that the inverse of any odd invertible function is odd.

#### Explanation:

The function $\tanh \left(x\right) \equiv \frac{{e}^{x} - {e}^{-} x}{{e}^{x} + {e}^{-} x}$ is obviously an odd function. So

$\tanh \left(- y\right) = - \tanh \left(y\right)$

Writing $\tanh \left(y\right) = x$, or equivalently $y = {\tanh}^{-} 1 \left(x\right)$, this equation becomes

$\tanh \left(- {\tanh}^{-} 1 \left(x\right)\right) = - x$

This implies

$- {\tanh}^{-} 1 \left(x\right) = {\tanh}^{-} 1 \left(- x\right)$

(where we have used ${\tanh}^{-} 1 \left(\tanh \left(x\right)\right) = x$)

So, ${\tanh}^{-} 1 \left(- x\right) = - {\tanh}^{-} 1 \left(x\right)$ - and thus $\tanh \left(x\right)$ is an odd function.

Apr 19, 2018

See the explanation below

#### Explanation:

The logarithmic form of the function

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

Substitute each $x$ by $- x$

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

Using properties of logarithmic functions

$\textcolor{g r e e n}{\ln \left(\frac{a}{b}\right) = \ln a - \ln b}$

$= \frac{1}{2} \left(\ln \left(1 - x\right) - \ln \left(1 + x\right)\right)$

take $- 1$ as a common factor

$= - \frac{1}{2} \left(\ln \left(1 + x\right) - \ln \left(1 - x\right)\right)$

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

$f \left(- x\right) = - f \left(x\right)$

$f \left(x\right)$ is an odd function.