Wells Adkins
08/31/2023 · Middle School
\( f(x)=x^{2}-x-1 \). On pose \( x_{0}=2 \). a. Résoudre l'équation \( x^{2}-x-1=0 \). On note \( \Phi \) la solution positive de cette équation. b. Montrer que pour tout entier naturel \( n \), on a: \( x_{n+1}=\frac{x_{n}^{2}+1}{2 x_{n}-1} \). c. Soit \( g \) la fonction définie sur \( [1 ;+\infty[ \) par \( g(x)=\frac{x^{2}+1}{2 x-1} \). Étudier les variations de la fonction \( g \).
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The positive solution of the equation \( x^2 - x - 1 = 0 \) is \( \Phi = \frac{1 + \sqrt{5}}{2} \). The relation \( x_{n+1} = \frac{x_n^2 + 1}{2x_n - 1} \) holds for all natural numbers \( n \). The function \( g(x) \) is increasing on the interval \( [1; +\infty[ \).
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