let-f-x-arctan-x-x-1-find-f-1-x-

Question Number 57411 by Abdo msup. last updated on 03/Apr/19

l e t f (x) = a r c t a n (\sqrt{x} + \sqrt{x + 1})

f i n d f^{- 1} (x) .

Commented by maxmathsup by imad last updated on 04/Apr/19

f i s d e f i n e d o n [0, + \infty [w e h a v e f (x) = y \Leftrightarrow x = f^{- 1} (y) a n d

f (x) = y \Rightarrow a r c t a n (\sqrt{x + 1} + \sqrt{x}) = y \Rightarrow \sqrt{x + 1} + \sqrt{x} = t a n y \Rightarrow

x + 1 + 2 \sqrt{x + 1} \sqrt{x} + x = {t a n}^{2} y \Rightarrow 2 x + 1 + 2 \sqrt{x + 1} \sqrt{x} = {t a n}^{2} y \Rightarrow

(2 x + 1 - {t a n}^{2} y) = 2 \sqrt{x + 1} \sqrt{x} \Rightarrow {(2 x + 1 - {t a n}^{2} y)}^{2} = 4 x (x + 1) \Rightarrow

{(2 x + 1)}^{2} - 2 (2 x + 1) {t a n}^{2} y + {t a n}^{4} y = 4 x^{2} + 4 x \Rightarrow

4 x^{2} + 4 x + 1 - 2 (2 x + 1) {t a n}^{2} y + {t a n}^{4} y = 4 x^{2} + 4 x \Rightarrow

1 + {t a n}^{4} y - 2 (2 x + 1) {t a n}^{2} y = 0 \Rightarrow 2 (2 x + 1) {t a n}^{2} y = 1 + {t a n}^{4} y \Rightarrow

2 (2 x + 1) = \frac{1 + {t a n}^{4} y}{{t a n}^{2} y} \Rightarrow 2 x + 1 = \frac{1 + {t a n}^{4} y}{2 {t a n}^{2} y} \Rightarrow 2 x = \frac{1 + {t a n}^{4} y}{2 {t a n}^{2} y} - 1 = \frac{{({t a n}^{2} y - 1)}^{2}}{2 {t a n}^{2} y}

\Rightarrow x = \frac{1}{4} \frac{{({t a n}^{2} y - 1)}^{2}}{{t a n}^{2} y} \Rightarrow f^{- 1} (x) = \frac{{({t a n}^{2} x - 1)}^{2}}{4 {t a n}^{2} x} .

Answered by kaivan.ahmadi last updated on 04/Apr/19

y = {t g}^{- 1} (\sqrt{x} + \sqrt{x + 1}) \Rightarrow t g y = \sqrt{x} + \sqrt{x + 1} \Rightarrow

{t g}^{2} y = x + x + 1 + 2 \sqrt{x (x + 1)} = 2 x + 1 + 2 \sqrt{x (x + 1)}

2 \sqrt{x} (t g y + \frac{1}{2 \sqrt{x}}) = 2 \sqrt{x} t g y + 1 \Rightarrow

(1 - 2 \sqrt{x}) t g y = 1 \Rightarrow 1 - 2 \sqrt{x} = \frac{1}{t g y} \Rightarrow

2 \sqrt{x} = 1 - \frac{1}{t g y} \Rightarrow \sqrt{x} = \frac{1}{2} - \frac{1}{2 t g y} \Rightarrow

x = {(\frac{1}{2} - \frac{1}{2 t g y})}^{2} \Rightarrow f^{- 1} (x) = \frac{1}{4} {(1 - \frac{1}{t g x})}^{2}