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Maxoucool
@Maxoucool
May 2019
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Bonjour, pour un exercice on me demande dresser le tableau de variation de f
f(x) = x³+2x+3
f'(x) =3x²+2
3x²+2 ≥ 0
3x²≥ -2
x²≥ -2/3
Et après ? il n'y pas de solution du coup ?
C'est juste croissant entre -infini et +infini ?
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ProfdeMaths1
Verified answer
f(x) = x³+2x+3
f'(x)=3x²+2
3x²+2>0 donc f'(x)>0
donc f est
strict croissante sur IR
2 votes
Thanks 1
Maxoucool
Merci !
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Verified answer
f(x) = x³+2x+3f'(x)=3x²+2
3x²+2>0 donc f'(x)>0
donc f est strict croissante sur IR