# What is the inverse of f(x)= -ln(arctan(x)) ?

Nov 13, 2015

${f}^{-} 1 \left(x\right) = \tan \left({e}^{-} x\right)$

#### Explanation:

A typical way of finding an inverse function is to set $y = f \left(x\right)$ and then solve for $x$ to obtain $x = {f}^{-} 1 \left(y\right)$

$y = - \ln \left(\arctan \left(x\right)\right)$

$\implies - y = \ln \left(\arctan \left(x\right)\right)$

$\implies {e}^{-} y = {e}^{\ln \left(\arctan \left(x\right)\right)} = \arctan \left(x\right)$ (by the definition of $\ln$)

$\implies \tan \left({e}^{-} y\right) = \tan \left(\arctan \left(x\right)\right) = x$ (by the definition of $\arctan$)

Thus we have ${f}^{-} 1 \left(x\right) = \tan \left({e}^{-} x\right)$

If we wish to confirm this via the definition ${f}^{-} 1 \left(f \left(x\right)\right) = f \left({f}^{-} 1 \left(x\right)\right) = x$
remember that $y = f \left(x\right)$ so we already have
${f}^{-} 1 \left(y\right) = {f}^{-} 1 \left(f \left(x\right)\right) = x$

For the reverse direction,

f(f^-1(x)) = -ln(arctan(tan(e^-x))

$\implies f \left({f}^{-} 1 \left(x\right)\right) = - \ln \left({e}^{-} x\right)$

$\implies f \left({f}^{-} 1 \left(x\right)\right) = - \left(- x \cdot \ln \left(e\right)\right) = - \left(- x \cdot 1\right)$

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