Question

prove that
$f(x) \: = \: (\frac{1}{x}) \: log \sqrt{x \: + \: \sqrt{ {x}^{2} \: + \: 1 } }$
is an even function. Please answer step by step solution​

Answer 1

Sorry but I don't know this function

Answer 2

Let y=log(x+(x2+1)−−−−−−−√ )

So. ey=x+(x2+1)−−−−−−−√

ie. ey−x=(x2+1)−−−−−−−√

(ey−x)2=x2+1

e2y−2xey+x2=x2+1

e2y−1=2xey

x=e2y−12eyex

So. f−1(x)=12(e(2x)−1)