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Question-201689




Question Number 201689 by cherokeesay last updated on 10/Dec/23
Answered by witcher3 last updated on 10/Dec/23
x^3 −6x^2 +12x−32=(x−2)^3 −24  x−2=y  ⇔((y+24))^(1/3) =y^3 −24  f(x)=(√(x+24))⇒f^− (y)=y^3 −24  conclusion withe f(x)=f^− (x)⇒f(x)=x  y^3 −24−y=0⇒(y−3)(x^2 +3x+8)=0  y=3;⇒x=5
$$\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} +\mathrm{12x}−\mathrm{32}=\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{3}} −\mathrm{24} \\ $$$$\mathrm{x}−\mathrm{2}=\mathrm{y} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{3}}]{\mathrm{y}+\mathrm{24}}=\mathrm{y}^{\mathrm{3}} −\mathrm{24} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}+\mathrm{24}}\Rightarrow\mathrm{f}^{−} \left(\mathrm{y}\right)=\mathrm{y}^{\mathrm{3}} −\mathrm{24} \\ $$$$\mathrm{conclusion}\:\mathrm{withe}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}^{−} \left(\mathrm{x}\right)\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x} \\ $$$$\mathrm{y}^{\mathrm{3}} −\mathrm{24}−\mathrm{y}=\mathrm{0}\Rightarrow\left(\mathrm{y}−\mathrm{3}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{3};\Rightarrow\mathrm{x}=\mathrm{5} \\ $$$$ \\ $$
Commented by cherokeesay last updated on 10/Dec/23
Nice ! thank you !
$${Nice}\:!\:{thank}\:{you}\:! \\ $$

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