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Question Number 32701 by caravan msup abdo. last updated on 31/Mar/18
let give f(x)= (x/( (√(x+1)))) 1)calculate f^(−1) (x)) 2) calculate (f^(−1) )^′ (x) .
$${let}\:{give}\:{f}\left({x}\right)=\:\frac{{x}}{\:\sqrt{{x}+\mathrm{1}}} \\ $$$$\left.\mathrm{1}\left.\right){calculate}\:{f}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)\:. \\ $$
Commented by Rio Mike last updated on 31/Mar/18
solution let f(x)=y y=(x/( (√(x+1)))) ((y(√(x+1)))/y)=(x/y) x+1=((x/y))^2 x=(x^2 /y^2 )−1 replacing y by x. f^(−1) (x)=((x^2_ /x^2 ))−1
$$\mathrm{solution} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{y} \\ $$$$\mathrm{y}=\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}+\mathrm{1}}} \\ $$$$\frac{\mathrm{y}\sqrt{\mathrm{x}+\mathrm{1}}}{\mathrm{y}}=\frac{\mathrm{x}}{\mathrm{y}} \\ $$$$\mathrm{x}+\mathrm{1}=\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{2}} \\ $$$$\mathrm{x}=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} }−\mathrm{1} \\ $$$$\mathrm{replacing}\:\mathrm{y}\:\mathrm{by}\:\mathrm{x}. \\ $$$$\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\left(\frac{\mathrm{x}^{\mathrm{2}_{} } }{\mathrm{x}^{\mathrm{2}} }\right)−\mathrm{1} \\ $$
Commented by MJS last updated on 31/Mar/18
you must exchange x and y each x must be replaced by y and each y must be replaced by x
$$\mathrm{you}\:\mathrm{must}\:\mathrm{exchange}\:{x}\:\mathrm{and}\:{y} \\ $$$$\boldsymbol{\mathrm{each}}\:{x}\:\mathrm{must}\:\mathrm{be}\:\mathrm{replaced}\:\mathrm{by}\:{y} \\ $$$$\mathrm{and}\:\boldsymbol{\mathrm{each}}\:{y}\:\mathrm{must}\:\mathrm{be}\:\mathrm{replaced}\:\mathrm{by}\:{x} \\ $$
Commented by abdo imad last updated on 01/Apr/18
let put f(x)=y ⇔x=f^(−1) (y) ⇒(x/( (√(x+1)))) =y ⇒x=y(√(x+1)) with conditiona x>−1 and (x/y)>0 ⇒x^2 =y^2 (x+1)=y^2 x +y^2 ⇒ x^2 −y^2 x −y^2 =0 equ.wth unknown x Δ = y^4 +4y^2 ≥0 ⇒ x_1 =((y^2 +(√(y^4 +4y^2 )))/2) x_2 = ((y^2 −(√(y^4 +4y^2 )))/2) but we must have x+1>0 let try with x_2 because its clear that x_1 +1>0 x_2 +1 = ((y^2 −(√(y^4 +4y^2 )))/2) +1 =((y^2 +2 −(√(y^4 +4y^2 )))/2) (y^2 +2)^(2 ) −(y^4 +4y^2 ) = 4 condition realized (x_2 /y) = ((y^2 −∣y∣(√(y^2 +4)))/(2y)) =y + ξ((√(4+y^2 ))/2) with ξ^2 =1 but the sine of (x_2 /y) is not constant so the so<ution of the equ.is x_1 and f^(−1) (x) = (1/2)(x^2 +(√(x^4 +4x^2 )) ) 2)(f^(−1) )^′ (x)= (1/2)(2x + ((4x^3 +8x)/(2(√(x^4 +4x^2 ))))) = x + ((2x^3 +4x)/( (√(x^4 +4x^2 )))) .
$${let}\:{put}\:{f}\left({x}\right)={y}\:\:\Leftrightarrow{x}={f}^{−\mathrm{1}} \left({y}\right)\:\Rightarrow\frac{{x}}{\:\sqrt{{x}+\mathrm{1}}}\:\:={y}\:\:\Rightarrow{x}={y}\sqrt{{x}+\mathrm{1}} \\ $$$${with}\:{conditiona}\:{x}>−\mathrm{1}\:{and}\:\frac{{x}}{{y}}>\mathrm{0}\:\:\Rightarrow{x}^{\mathrm{2}} \:={y}^{\mathrm{2}} \left({x}+\mathrm{1}\right)={y}^{\mathrm{2}} {x}\:+{y}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} {x}\:−{y}^{\mathrm{2}} =\mathrm{0}\:\:{equ}.{wth}\:{unknown}\:{x} \\ $$$$\Delta\:=\:{y}^{\mathrm{4}} \:\:+\mathrm{4}{y}^{\mathrm{2}} \:\geqslant\mathrm{0}\:\:\Rightarrow\:{x}_{\mathrm{1}} \:=\frac{{y}^{\mathrm{2}} \:+\sqrt{{y}^{\mathrm{4}} \:+\mathrm{4}{y}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\:\frac{{y}^{\mathrm{2}} \:−\sqrt{{y}^{\mathrm{4}} \:+\mathrm{4}{y}^{\mathrm{2}} }}{\mathrm{2}}\:{but}\:{we}\:{must}\:{have}\:{x}+\mathrm{1}>\mathrm{0}\:{let}\:{try} \\ $$$${with}\:{x}_{\mathrm{2}} \:\:{because}\:{its}\:{clear}\:{that}\:{x}_{\mathrm{1}} \:+\mathrm{1}>\mathrm{0} \\ $$$${x}_{\mathrm{2}} \:+\mathrm{1}\:=\:\frac{{y}^{\mathrm{2}} \:−\sqrt{{y}^{\mathrm{4}} \:+\mathrm{4}{y}^{\mathrm{2}} }}{\mathrm{2}}\:+\mathrm{1}\:\:=\frac{{y}^{\mathrm{2}} \:+\mathrm{2}\:−\sqrt{{y}^{\mathrm{4}} \:+\mathrm{4}{y}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\left({y}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{2}\:} \:−\left({y}^{\mathrm{4}} \:+\mathrm{4}{y}^{\mathrm{2}} \right)\:=\:\mathrm{4}\:{condition}\:{realized}\: \\ $$$$\frac{{x}_{\mathrm{2}} }{{y}}\:=\:\frac{{y}^{\mathrm{2}} \:−\mid{y}\mid\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}{y}}\:={y}\:+\:\xi\frac{\sqrt{\mathrm{4}+{y}^{\mathrm{2}} }}{\mathrm{2}}\:{with}\:\xi^{\mathrm{2}} \:=\mathrm{1}\:{but}\:{the}\:{sine} \\ $$$${of}\:\:\frac{{x}_{\mathrm{2}} }{{y}}\:{is}\:{not}\:{constant}\:\:{so}\:{the}\:{so}<{ution}\:{of}\:{the}\:{equ}.{is}\:{x}_{\mathrm{1}} \\ $$$${and}\:{f}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\sqrt{{x}^{\mathrm{4}} \:+\mathrm{4}{x}^{\mathrm{2}} }\:\right) \\ $$$$\left.\mathrm{2}\right)\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}\:+\:\frac{\mathrm{4}{x}^{\mathrm{3}} \:+\mathrm{8}{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{4}} \:+\mathrm{4}{x}^{\mathrm{2}} }}\right) \\ $$$$=\:{x}\:\:+\:\frac{\mathrm{2}{x}^{\mathrm{3}} \:+\mathrm{4}{x}}{\:\sqrt{{x}^{\mathrm{4}} \:+\mathrm{4}{x}^{\mathrm{2}} }}\:. \\ $$
Answered by MJS last updated on 31/Mar/18
x=(y/( (√(y+1)))) x(√(y+1))=y x^2 y+x^2 =y^2 y^2 −x^2 y−x^2 =0 y=(x^2 /2)±(√((x^4 /4)+x^2 )) y=(x^2 /2)±(√((x^2 /4)(x^2 +4))) y=f^(−1) (x)=(x^2 /2)±(x/2)(√(x^2 +4)) y′=((2x)/2)±((1/2)×(x^2 +4)^(1/2) +(x/2)×(1/2)×(x^2 +4)^(−(1/2)) ×2x) y′=x±((1/2)(x^2 +4)^(1/2) +(x^2 /2)(x^2 +4)^(−(1/2)) ) y′=x±(1/2)(x^2 +4)^(−(1/2)) ((x^2 +4)^1 +x^2 ) y′=(f^(−1) )′(x)=x±((x^2 +2)/( (√(x^2 +4))))
$${x}=\frac{{y}}{\:\sqrt{{y}+\mathrm{1}}} \\ $$$${x}\sqrt{{y}+\mathrm{1}}={y} \\ $$$${x}^{\mathrm{2}} {y}+{x}^{\mathrm{2}} ={y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\pm\sqrt{\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+{x}^{\mathrm{2}} } \\ $$$${y}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\pm\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${y}={f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\pm\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$$ \\ $$$${y}'=\frac{\mathrm{2}{x}}{\mathrm{2}}\pm\left(\frac{\mathrm{1}}{\mathrm{2}}×\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\frac{{x}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}×\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{2}{x}\right) \\ $$$${y}'={x}\pm\left(\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$${y}'={x}\pm\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{1}} +{x}^{\mathrm{2}} \right) \\ $$$${y}'=\left({f}^{−\mathrm{1}} \right)'\left({x}\right)={x}\pm\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$

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