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Question Number 32283 by abdo imad last updated on 22/Mar/18

let give f(x)=x+2 −(√(x+1)) 1) find f^(−1) (x) inverse of f(x) 2) calculate (f^(−1) )^′ (x) .

$${let}\:{give}\:{f}\left({x}\right)={x}+\mathrm{2}\:−\sqrt{{x}+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}^{−\mathrm{1}} \left({x}\right)\:{inverse}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)\:. \\ $$

Commented by abdo imad last updated on 24/Mar/18

D_f =[−1,+∞[ f(x)=y ⇔ y=x+2 −(√(x+1)) ⇔ (x+2 −y)^2 =x+1 ⇒ (x+2)^2 −2y(x+2) +y^2 −x−1=0 ⇒ x^2 +4x +4 −2yx −4y +y^2 −x−1 =0⇒ x^2 +(3−2y)x +y^2 −4y +3 =0 Δ = (3−2y)^2 −4(y^2 −4y +3)= =9 −12y +4y^2 −4y^2 +16y −12 =4y −3 x_1 = ((−3+2y +(√( 4y−3)))/2) and x_2 =((−3 +2y −(√(4h−3)))/2) we must have x+2−y ≥0 let verify x_2 +2 −y =((−3 +2y −(√(4y −3)))/2) +2−y =((−3 +2y −(√(4y−3)) +4−2y)/2) = ((1−y −(√(4y−3)))/2) <0 so the soution is x_1 and f^(−1) (x) =((2x +(√(4x−3)) −3)/2) f^(−1) (x) = x +(1/2)(√(4x−3)) −(3/2) ⇒ (f^(−1) )^′ (x) = 1+ (1/2) (4/(2(√(4x−3)))) = 1+ (1/(√(4x−3))) . with x>(3/4) .

$${D}_{{f}} =\left[−\mathrm{1},+\infty\left[\:\:{f}\left({x}\right)={y}\:\Leftrightarrow\:{y}={x}+\mathrm{2}\:−\sqrt{{x}+\mathrm{1}}\:\Leftrightarrow\right.\right. \\ $$$$\left({x}+\mathrm{2}\:−{y}\right)^{\mathrm{2}} \:={x}+\mathrm{1}\:\Rightarrow\:\left({x}+\mathrm{2}\right)^{\mathrm{2}} \:−\mathrm{2}{y}\left({x}+\mathrm{2}\right)\:+{y}^{\mathrm{2}} \:−{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:+\mathrm{4}\:−\mathrm{2}{yx}\:−\mathrm{4}{y}\:+{y}^{\mathrm{2}} \:−{x}−\mathrm{1}\:=\mathrm{0}\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+\left(\mathrm{3}−\mathrm{2}{y}\right){x}\:+{y}^{\mathrm{2}} \:−\mathrm{4}{y}\:+\mathrm{3}\:=\mathrm{0} \\ $$$$\Delta\:=\:\left(\mathrm{3}−\mathrm{2}{y}\right)^{\mathrm{2}} \:−\mathrm{4}\left({y}^{\mathrm{2}} \:−\mathrm{4}{y}\:+\mathrm{3}\right)= \\ $$$$=\mathrm{9}\:−\mathrm{12}{y}\:+\mathrm{4}{y}^{\mathrm{2}} \:−\mathrm{4}{y}^{\mathrm{2}} \:+\mathrm{16}{y}\:−\mathrm{12}\:=\mathrm{4}{y}\:−\mathrm{3} \\ $$$${x}_{\mathrm{1}} =\:\frac{−\mathrm{3}+\mathrm{2}{y}\:+\sqrt{\:\mathrm{4}{y}−\mathrm{3}}}{\mathrm{2}}\:{and}\:\:{x}_{\mathrm{2}} =\frac{−\mathrm{3}\:+\mathrm{2}{y}\:−\sqrt{\mathrm{4}{h}−\mathrm{3}}}{\mathrm{2}} \\ $$$${we}\:{must}\:{have}\:\:{x}+\mathrm{2}−{y}\:\geqslant\mathrm{0}\:{let}\:{verify} \\ $$$${x}_{\mathrm{2}} \:+\mathrm{2}\:−{y}\:=\frac{−\mathrm{3}\:+\mathrm{2}{y}\:−\sqrt{\mathrm{4}{y}\:−\mathrm{3}}}{\mathrm{2}}\:+\mathrm{2}−{y} \\ $$$$=\frac{−\mathrm{3}\:+\mathrm{2}{y}\:−\sqrt{\mathrm{4}{y}−\mathrm{3}}\:+\mathrm{4}−\mathrm{2}{y}}{\mathrm{2}}\:=\:\frac{\mathrm{1}−{y}\:−\sqrt{\mathrm{4}{y}−\mathrm{3}}}{\mathrm{2}}\:<\mathrm{0}\:{so}\:{the} \\ $$$${soution}\:{is}\:{x}_{\mathrm{1}} \:\:{and}\:{f}^{−\mathrm{1}} \left({x}\right)\:=\frac{\mathrm{2}{x}\:+\sqrt{\mathrm{4}{x}−\mathrm{3}}\:−\mathrm{3}}{\mathrm{2}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}{x}−\mathrm{3}}\:−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$$\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)\:=\:\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{4}{x}−\mathrm{3}}}\:=\:\mathrm{1}+\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{4}{x}−\mathrm{3}}}\:.\:{with}\:{x}>\frac{\mathrm{3}}{\mathrm{4}}\:. \\ $$

Answered by MJS last updated on 22/Mar/18

x=y+2−(√(y+1)) (√(y+1))=−x+y+2 y+1=y^2 +2(2−x)y+(2−x)^2 y^2 +(3−2x)y+(x^2 −4x+3)=0 p=3−2x; q=(x^2 −4x+3) y=−(p/2)±((√(p^2 −4q))/2) y=x−(3/2)±((√(4x−3))/2) y′=1±(((1/2)(4x−3)^(−(1/2)) ×4)/2) y′=1±(1/(√(4x−3)))

$${x}={y}+\mathrm{2}−\sqrt{{y}+\mathrm{1}} \\ $$$$\sqrt{{y}+\mathrm{1}}=−{x}+{y}+\mathrm{2} \\ $$$${y}+\mathrm{1}={y}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}−{x}\right){y}+\left(\mathrm{2}−{x}\right)^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{2}{x}\right){y}+\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$${p}=\mathrm{3}−\mathrm{2}{x};\:{q}=\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right) \\ $$$${y}=−\frac{{p}}{\mathrm{2}}\pm\frac{\sqrt{{p}^{\mathrm{2}} −\mathrm{4}{q}}}{\mathrm{2}} \\ $$$${y}={x}−\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}{x}−\mathrm{3}}}{\mathrm{2}} \\ $$$${y}'=\mathrm{1}\pm\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{x}−\mathrm{3}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{4}}{\mathrm{2}} \\ $$$${y}'=\mathrm{1}\pm\frac{\mathrm{1}}{\sqrt{\mathrm{4}{x}−\mathrm{3}}} \\ $$

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