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Question Number 185090 by SEKRET last updated on 16/Jan/23
x^3 +1 = 2((2x−1))^(1/3) x=?
$$\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{1}\:=\:\:\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=? \\ $$$$ \\ $$$$ \\ $$
Commented by Frix last updated on 16/Jan/23
Obviously x=1 is one of the solutions
$$\mathrm{Obviously}\:{x}=\mathrm{1}\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solutions} \\ $$
Commented by Frix last updated on 16/Jan/23
If we stay in R ⇒ ((−r))^(1/3) =−(r)^(1/3) we get x_1 =−((1+(√5))/2)=−ϕ x_2 =−((1−(√5))/2)=(1/ϕ) x_3 =1
$$\mathrm{If}\:\mathrm{we}\:\mathrm{stay}\:\mathrm{in}\:\mathbb{R}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=−\varphi \\ $$$${x}_{\mathrm{2}} =−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}=\frac{\mathrm{1}}{\varphi} \\ $$$${x}_{\mathrm{3}} =\mathrm{1} \\ $$
Answered by floor(10²Eta[1]) last updated on 16/Jan/23
f(x)=((x^3 +1)/2)⇒f^(−1) (x)=((2x−1))^(1/3) so we want to solve: f(x)=f^(−1) (x)⇒f(f(x))=x note that f(x) is an increasing function because f′(x)=((3x^2 )/2)>0, ∀ x∈R^+ ⇒ { ((if f(x)>x⇒f(f(x))>f(x)>x)),((if f(x)<x⇒f(f(x))<f(x)<x)) :} but we want f(f(x))=x⇒f(x)=x ((x^3 +1)/2)=x⇒x^3 −2x+1=0 x=1 is a solution. x^3 −2x+1=(x−1)(x^2 +x−1)=0 ⇒x=((−1±(√5))/2)
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\sqrt[{\mathrm{3}}]{\mathrm{2x}−\mathrm{1}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{solve}: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\Rightarrow\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x} \\ $$$$ \\ $$$$\mathrm{note}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{increasing}\:\mathrm{function} \\ $$$$\mathrm{because}\:\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{2}}>\mathrm{0},\:\forall\:\mathrm{x}\in\mathbb{R}^{+} \\ $$$$\Rightarrow\begin{cases}{\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)>\mathrm{x}\Rightarrow\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)>\mathrm{f}\left(\mathrm{x}\right)>\mathrm{x}}\\{\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)<\mathrm{x}\Rightarrow\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)<\mathrm{f}\left(\mathrm{x}\right)<\mathrm{x}}\end{cases} \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{want}\:\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x} \\ $$$$ \\ $$$$\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}{\mathrm{2}}=\mathrm{x}\Rightarrow\mathrm{x}^{\mathrm{3}} −\mathrm{2x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{2x}+\mathrm{1}=\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\: \\ $$

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