Απαντήσει&I

1 ξ β f ξ f β f β f ξ άρα g 1 οπότε g 1

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1 ξ β f ξ f β f β f ξ 0 άρα   G 1 0 Οπότε     G 1 G 3 0 Ισχύουν οι προϋποθέσεις του Θεωρήματος Bolzano άρα η εξίσωση:        3 5 G x 0 x 3 F β 1 β f β x x 1 β 1 x 1 0       3 5 F β 1 β f β β 1 x 1 x 0 x 1 x 3 έχει μια τουλάχιστον ρίζα στο (1,3) Δ4 ) Για το   2 x x t Ι x f dx x έχω t θ t x άρα dt xd θ Οπότε       x x 1 1 Ι x xf θ dθ xf t dt H ανισοϊσότητα ισοδύναμα γράφεται για x 0           2 x x x x x x x 1 1 1 1 1 t f dx tf t dt xf t dt tf t dt xf t dt tf t dt 0 x        x x 1 1 xf t tf t dt 0 x t f t dt 0 (4) Για x 1 η (4) προφανώς ισχύει Για x 1 τότε t 1,x άρα t x x t 0 και     f γν. φθιν. t 1 f t f 1 0 Οπότε    x t f t 0 άρα    x 1 x t f t dt 0 Για 0 x 1 τότε t x,1 άρα x t 1 x t 0 και     f γν. φθιν. t 1 f t f 1 0 Οπότε    x t f t 0 άρα       1 x x 1 x t f t dt 0 x t f t dt 0 Έτσι λοιπόν για κάθε x 0 είναι    x 1 x t f t dt 0 t x x 2 θ 1 x Νέα άκρα
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