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Question Number 57411 by Abdo msup. last updated on 03/Apr/19
let f(x)=arctan((√x)+(√(x+1))) find f^(−1) (x) .
$${let}\:{f}\left({x}\right)={arctan}\left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right) \\ $$$${find}\:{f}^{−\mathrm{1}} \left({x}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 04/Apr/19
f is defined on [0,+∞[ we have f(x)=y ⇔x =f^(−1) (y) and f(x)=y ⇒arctan((√(x+1))+(√x)) =y ⇒(√(x+1))+(√x)=tany ⇒ x+1 +2(√(x+1))(√x)+x =tan^2 y ⇒2x+1 +2(√(x+1))(√x)=tan^2 y ⇒ (2x+1−tan^2 y)=2(√(x+1))(√x) ⇒(2x+1−tan^2 y)^2 =4x(x+1) ⇒ (2x+1)^2 −2(2x+1)tan^2 y +tan^4 y = 4x^2 +4x ⇒ 4x^2 +4x +1 −2(2x+1)tan^2 y +tan^4 y =4x^2 +4x ⇒ 1+tan^4 y −2(2x+1)tan^2 y =0 ⇒2(2x+1)tan^2 y = 1+tan^4 y ⇒ 2(2x+1) =((1+tan^4 y)/(tan^2 y)) ⇒2x+1 =((1+tan^4 y)/(2tan^2 y)) ⇒2x =((1+tan^4 y)/(2tan^2 y)) −1 =(((tan^2 y−1)^2 )/(2tan^2 y)) ⇒x =(1/4) (((tan^2 y−1)^2 )/(tan^2 y)) ⇒ f^(−1) (x) =(((tan^2 x−1)^2 )/(4tan^2 x)) .
$${f}\:{is}\:{defined}\:{on}\:\left[\mathrm{0},+\infty\left[\:\:\:{we}\:{have}\:\:{f}\left({x}\right)={y}\:\Leftrightarrow{x}\:={f}^{−\mathrm{1}} \left({y}\right)\:{and}\:\right.\right. \\ $$$${f}\left({x}\right)={y}\:\Rightarrow{arctan}\left(\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}\right)\:={y}\:\Rightarrow\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}={tany}\:\Rightarrow \\ $$$${x}+\mathrm{1}\:+\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{x}}+{x}\:={tan}^{\mathrm{2}} {y}\:\Rightarrow\mathrm{2}{x}+\mathrm{1}\:+\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{x}}={tan}^{\mathrm{2}} {y}\:\Rightarrow \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}−{tan}^{\mathrm{2}} {y}\right)=\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{x}}\:\Rightarrow\left(\mathrm{2}{x}+\mathrm{1}−{tan}^{\mathrm{2}} {y}\right)^{\mathrm{2}} =\mathrm{4}{x}\left({x}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right){tan}^{\mathrm{2}} {y}\:+{tan}^{\mathrm{4}} {y}\:=\:\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:\Rightarrow \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:+\mathrm{1}\:−\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right){tan}^{\mathrm{2}} {y}\:+{tan}^{\mathrm{4}} {y}\:=\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:\Rightarrow \\ $$$$\mathrm{1}+{tan}^{\mathrm{4}} {y}\:−\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right){tan}^{\mathrm{2}} {y}\:=\mathrm{0}\:\Rightarrow\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right){tan}^{\mathrm{2}} {y}\:=\:\mathrm{1}+{tan}^{\mathrm{4}} {y}\:\Rightarrow \\ $$$$\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)\:=\frac{\mathrm{1}+{tan}^{\mathrm{4}} {y}}{{tan}^{\mathrm{2}} {y}}\:\Rightarrow\mathrm{2}{x}+\mathrm{1}\:=\frac{\mathrm{1}+{tan}^{\mathrm{4}} {y}}{\mathrm{2}{tan}^{\mathrm{2}} {y}}\:\Rightarrow\mathrm{2}{x}\:=\frac{\mathrm{1}+{tan}^{\mathrm{4}} {y}}{\mathrm{2}{tan}^{\mathrm{2}} {y}}\:−\mathrm{1}\:=\frac{\left({tan}^{\mathrm{2}} {y}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{tan}^{\mathrm{2}} {y}} \\ $$$$\Rightarrow{x}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\left({tan}^{\mathrm{2}} {y}−\mathrm{1}\right)^{\mathrm{2}} }{{tan}^{\mathrm{2}} {y}}\:\Rightarrow\:{f}^{−\mathrm{1}} \left({x}\right)\:=\frac{\left({tan}^{\mathrm{2}} {x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}{tan}^{\mathrm{2}} {x}}\:. \\ $$
Answered by kaivan.ahmadi last updated on 04/Apr/19
y=tg^(−1) ((√x)+(√(x+1)))⇒tgy=(√x)+(√(x+1))⇒ tg^2 y=x+x+1+2(√(x(x+1)))=2x+1+2(√(x(x+1))) 2(√x)(tgy+(1/(2(√x))))=2(√x)tgy+1⇒ (1−2(√x))tgy=1⇒1−2(√x)=(1/(tgy))⇒ 2(√x)=1−(1/(tgy))⇒(√x)=(1/2)−(1/(2tgy))⇒ x=((1/2)−(1/(2tgy)))^2 ⇒f^(−1) (x)=(1/4)(1−(1/(tgx)))^2
$${y}={tg}^{−\mathrm{1}} \left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right)\Rightarrow{tgy}=\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\Rightarrow \\ $$$${tg}^{\mathrm{2}} {y}={x}+{x}+\mathrm{1}+\mathrm{2}\sqrt{{x}\left({x}+\mathrm{1}\right)}=\mathrm{2}{x}+\mathrm{1}+\mathrm{2}\sqrt{{x}\left({x}+\mathrm{1}\right)} \\ $$$$\mathrm{2}\sqrt{{x}}\left({tgy}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\right)=\mathrm{2}\sqrt{{x}}{tgy}+\mathrm{1}\Rightarrow \\ $$$$\left(\mathrm{1}−\mathrm{2}\sqrt{{x}}\right){tgy}=\mathrm{1}\Rightarrow\mathrm{1}−\mathrm{2}\sqrt{{x}}=\frac{\mathrm{1}}{{tgy}}\Rightarrow \\ $$$$\mathrm{2}\sqrt{{x}}=\mathrm{1}−\frac{\mathrm{1}}{{tgy}}\Rightarrow\sqrt{{x}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}{tgy}}\Rightarrow \\ $$$${x}=\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}{tgy}}\right)^{\mathrm{2}} \Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\frac{\mathrm{1}}{{tgx}}\right)^{\mathrm{2}} \\ $$

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