Given-f-x-1-x-2-x-x-1-find-f-x-

Question Number 139029 by bramlexs22 last updated on 21/Apr/21

Given f (x + \sqrt{1 + x^{2}}) = \frac{x}{x + 1}

find f (x) = ?

Answered by EDWIN88 last updated on 21/Apr/21

u = x + \sqrt{1 + x^{2}}

u^{2} - 2 ux + x^{2} = 1 + x^{2}

x = \frac{u^{2} - 1}{2 u} \Rightarrow f (u) = \frac{\frac{u^{2} - 1}{2 u}}{\frac{u^{2} - 1}{2 u} + 1} = \frac{u^{2} - 1}{u^{2} + 2 u - 1}

then f (x) = \frac{x^{2} - 1}{x^{2} + 2 x - 1}

Answered by mathmax by abdo last updated on 22/Apr/21

let x = sht \Rightarrow t = argshx = \log (x + \sqrt{1 + x^{2}})

f (x + \sqrt{1 + x^{2}}) = f (sht + cht) = \frac{sht}{1 + sht} \Rightarrow f (e^{t}) = \frac{\frac{e^{t} - e^{- t}}{2}}{1 + \frac{e^{t} - e^{- t}}{2}}

= \frac{e^{t} - e^{- t}}{2 + e^{t} - e^{- t}} let e^{t} = x \Rightarrow f (x) = \frac{x - x^{- 1}}{2 + x - x^{-}} = \frac{x^{2} - 1}{2 x + x^{2} - 1} \Rightarrow

f (x) = \frac{x^{2} - 1}{x^{2} + 2 x - 1}