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celimegharbi
@celimegharbi
January 2021
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Bonjour j'ai besoin d'aide a ces question
merci d'avance!!!!
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anylor
Bonjour
voici la réponse en fichier joint
1 votes
Thanks 1
celimegharbi
MERCI BEAUCOUP :)
syogier
F(1) = -2 (1)² +1 -1 = -2
f(1+h) = -2(1+h)² +( 1+h) -1 = -2 (1²+h²+2h) +1+h-1 = -2-2h²-4h+h
=-2h²-3h-2
(f(1+h) -f(1)) /h =(-2h²-3h-2 -(-2)) /h = (-2h²-3h)/h = h(-2h-3) /h = -2h -3
f'(1) = lim (-2h-3) quand h tend vers zéro f'(1) = -3
l'équation de la tangente au point d'abscisse 1 est y = f'(1) (x-1) +f(1) . Je te laisse calculer
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celimegharbi
MERCI INFINIMENT !
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BONSOIR DM DE MATHS URGENT JAI BESOIN DAIDE SVP!!!!! MERCI DAVANCE..
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voici la réponse en fichier joint
f(1+h) = -2(1+h)² +( 1+h) -1 = -2 (1²+h²+2h) +1+h-1 = -2-2h²-4h+h
=-2h²-3h-2
(f(1+h) -f(1)) /h =(-2h²-3h-2 -(-2)) /h = (-2h²-3h)/h = h(-2h-3) /h = -2h -3
f'(1) = lim (-2h-3) quand h tend vers zéro f'(1) = -3
l'équation de la tangente au point d'abscisse 1 est y = f'(1) (x-1) +f(1) . Je te laisse calculer