x-3-1-2-2x-1-1-3-x-

Question Number 185090 by SEKRET last updated on 16/Jan/23

x^{3} + 1 = 2 \sqrt[3]{2 x - 1}

x = ?

Commented by Frix last updated on 16/Jan/23

Obviously x = 1 is one of the solutions

Commented by Frix last updated on 16/Jan/23

If we stay in R \Rightarrow \sqrt[3]{- r} = - \sqrt[3]{r} we get

x_{1} = - \frac{1 + \sqrt{5}}{2} = - φ

x_{2} = - \frac{1 - \sqrt{5}}{2} = \frac{1}{φ}

x_{3} = 1

Answered by floor(10²Eta[1]) last updated on 16/Jan/23

f (x) = \frac{x^{3} + 1}{2} \Rightarrow f^{- 1} (x) = \sqrt[3]{2 x - 1}

so we want to solve :

f (x) = f^{- 1} (x) \Rightarrow f (f (x)) = x

note that f (x) is an increasing function

because f^{'} (x) = \frac{{3 x}^{2}}{2} > 0, \forall x \in R^{+}

\Rightarrow {\begin{cases} if f (x) > x \Rightarrow f (f (x)) > f (x) > x \\ if f (x) < x \Rightarrow f (f (x)) < f (x) < x \end{cases}

but we want f (f (x)) = x \Rightarrow f (x) = x

\frac{x^{3} + 1}{2} = x \Rightarrow x^{3} - 2 x + 1 = 0

x = 1 is a solution .

x^{3} - 2 x + 1 = (x - 1) (x^{2} + x - 1) = 0

\Rightarrow x = \frac{- 1 \pm \sqrt{5}}{2}