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

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

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Answer 1 · 2018-04-19T12:51:10

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

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

$tanh (- y) = - tanh (y)$ $tanh(-y) = -tanh(y)$

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

$tanh (- {tanh}^{- 1} (x)) = - x$ $tanh(-tanh^-1(x))=-x$

This implies

$- {tanh}^{- 1} (x) = {tanh}^{- 1} (- x)$ $-tanh^-1(x) =tanh^-1(-x)$

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

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

Answer 2 · 2018-04-19T12:56:19

See the explanation below

The logarithmic form of the function

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

Substitute each $x$ $x$ by $- x$ $-x$

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

Using properties of logarithmic functions

$ln (\frac{a}{b}) = ln a - ln b$ $color(green) (ln(a/b)=lna-lnb)$

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

take $- 1$ $-1$ as a common factor

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

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

$f (- x) = - f (x)$ $f(-x)=-f(x)$

$f (x)$ $f(x)$ is an odd function.