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Marius963
Pe: 2014-01-30T10:46:57+02:00 In: In:

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Pentru fiecare n∈N, n ≥ 3 , se consideră funcţia Fn :[0, ∞)→R, Fn(x)=x^n-nx+1.

Să se arate că ecuaţia Fn (x) = 0, x > 0 are exact două rădăcini an ∈(0,1) şi bn ∈(1,∞) .

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2 raspunsuri

  1. DD
    profesor


    Functia Fn(x) este un polinom continu care isi schimba semnul in intervalul (0,1)-(Fn(0)=1 si Fn(1)=-(n-1))unde trebue sa aiba o radacina si in intervalul (1 , +inf) -(lim(x->inf.)Fn(x)->+inf,) unde mai trebuesa aibe inca o radacina

  2. gunty
    maestru (V)

        \[ \begin{array}{l} Fie\;\left( {x_k } \right)_{k \ge 0} \;un\;sir\;cu\;x_k \ge 0\;si\;\lim x_k = x_0 \in \left[ {0, + \infty } \right). \\ Avem\; {\lim }\limits_{k \to \infty } f_n \left( {x_k } \right) = {\lim }\limits_{k \to \infty } x_k ^n - {\lim }\limits_{k \to \infty } \left( {n \cdot x_k } \right) + 1 = x_0 ^n - nx_0 + 1 = f_n \left( {x_0 } \right). \\ Conform\;criteriului\;Heine,\;f_n \;este\;continua\;in\;orice\;punct\;x_0 ,\;adica\;f_n \;este\;functie\;continua,\;deci\;are\;proprietatea\;Darboux. \\ \end{array} \]

        \[ \begin{array}{l} Fie\;0 < x_1 < x_2 < 1. \\ Atunci\;f_n \left( {x_1 } \right) > f_n \left( {x_2 } \right) \Leftrightarrow x_1 ^n - nx_1 + 1 > x_2 ^n - nx_2 + 1 \Leftrightarrow x_1 ^n - x_2 ^n > n\left( {x_1 - x_2 } \right) \Leftrightarrow \left( {x_1 - x_2 } \right)\left( {x_1 ^{n - 1} + x_1 ^{n - 2} x_2 + ... + x_1 x_2 ^{n - 2} + x_2 ^{n - 1} } \right) > n\left( {x_1 - x_2 } \right) \Leftrightarrow \\ \Leftrightarrow x_1 ^{n - 1} + x_1 ^{n - 2} x_2 + ... + x_1 x_2 ^{n - 2} + x_2 ^{n - 1} < n \Leftrightarrow \left( {x_1 ^{n - 1} - 1} \right) + \left( {x_1 ^{n - 2} x_2 - 1} \right) + ... + \left( {x_1 x_2 ^{n - 2} - 1} \right) + \left( {x_2 ^{n - 1} - 1} \right) < 0,\;adevarat! \\ Deci\;functia\;este\;strict\;descrescatoare\;pe\;intervalul\;\left( {0,1} \right). \\ \end{array} \]

        \[ \begin{array}{l} Fie\;1 < x_1 < x_2 . \\ Atunci\;f_n \left( {x_1 } \right) < f_n \left( {x_2 } \right) \Leftrightarrow x_1 ^n - nx_1 + 1 < x_2 ^n - nx_2 + 1 \Leftrightarrow x_1 ^n - x_2 ^n < n\left( {x_1 - x_2 } \right) \Leftrightarrow \left( {x_1 - x_2 } \right)\left( {x_1 ^{n - 1} + x_1 ^{n - 2} x_2 + ... + x_1 x_2 ^{n - 2} + x_2 ^{n - 1} } \right) < n\left( {x_1 - x_2 } \right) \Leftrightarrow \\ \Leftrightarrow x_1 ^{n - 1} + x_1 ^{n - 2} x_2 + ... + x_1 x_2 ^{n - 2} + x_2 ^{n - 1} > n \Leftrightarrow \left( {x_1 ^{n - 1} - 1} \right) + \left( {x_1 ^{n - 2} x_2 - 1} \right) + ... + \left( {x_1 x_2 ^{n - 2} - 1} \right) + \left( {x_2 ^{n - 1} - 1} \right) > 0,\;adevarat! \\ Deci\;functia\;este\;strict\;crescatoare\;pe\;intervalul\;\left( {1, + \infty } \right). \\ \end{array} \]

        \[ \begin{array}{l} f_n \left( 0 \right) = 1 > 0\;si\;f_n \left( 1 \right) = - n < 0,\;iar\;f_n \;are\;proprietatea\;Darboux.\;Deducem\;ca\;exista\;a_n \in \left( {0,1} \right)\;astfel\;incat\;f_n \left( {a_n } \right) = 0. \\ Din\;monotonie\;rezulta\;ca\;exista\;doar\;un\;singur\;a_n \in \left( {0,1} \right)\;pentru\;care\;f_n \left( {a_n } \right) = 0,\;adica\;ecuatia\;f_n \left( x \right) = 0\;are\;solutie\;unica\;pe\;intervalul\;\left( {0,1} \right). \\ \end{array} \]

        \[ \begin{array}{l} f_n \left( 1 \right) = - n < 0\;si\;f_n \left( n \right) = n^n - n^2 + 1 > n^2 - n^2 + 1 = 1 > 0,\;iar\;f_n \;are\;proprietatea\;Darboux.\;Deducem\;ca\;exista\;b_n \in \left( {1,n} \right)\;astfel\;incat\;f_n \left( {b_n } \right) = 0. \\ Din\;monotonie\;rezulta\;ca\;exista\;doar\;un\;singur\;b_n \in \left( {0,1} \right)\;pentru\;care\;f_n \left( {b_n } \right) = 0,\;adica\;ecuatia\;f_n \left( x \right) = 0\;are\;solutie\;unica\;pe\;intervalul\;\left( {1,n} \right). \\ Deoarece\;f_n \left( n \right) > 0\;si\;f_n \;este\;strict\;crescatoare\;pe\;\left[ {n, + \infty } \right) \subset \left( {1, + \infty } \right),\;deducem\;ca\;ecuatia\;f_n \left( x \right) = 0\;nu\;are\;solutii\;pe\;\left[ {n, + \infty } \right). \\ Asadar\;ecuatia\;f_n \left( x \right) = 0\;are\;exact\;o\;solutie\;pe\;intervalul\;\left( {1, + \infty } \right). \\ \end{array} \]

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