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Chris51
@Chris51
April 2019
1
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Etude des signes:
d) f(x) = -4x-1
e) f(x) = (x-1)(x-2) + (x-1)(4x-4)
MERCI D'AVANCE !!!!!!!!
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raymrich
D)
x = -1/4 ⇒ f(x) = 0
-4x-1>0 ⇔ -4x>1 ⇔ x<-1/4; d'où x<-1/4 ⇒ f(x)>0
-4x-1<0 ⇔ -4x<1 ⇔ x>-1/4; d'où x>-1/4 ⇒ f(x)<0
e)
(x-1)(x-2)+(x-1)(4x-4) = (x-1)(x-2+4x-4) = (x-1)(5x-6)
x = 1 ou x = 6/5 ⇒ f(x) = 0
x<1 ou x>6/5 ⇒ f(x)>0
1<x<6/5 ⇒ f(x)<0
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x = -1/4 ⇒ f(x) = 0
-4x-1>0 ⇔ -4x>1 ⇔ x<-1/4; d'où x<-1/4 ⇒ f(x)>0
-4x-1<0 ⇔ -4x<1 ⇔ x>-1/4; d'où x>-1/4 ⇒ f(x)<0
e)
(x-1)(x-2)+(x-1)(4x-4) = (x-1)(x-2+4x-4) = (x-1)(5x-6)
x = 1 ou x = 6/5 ⇒ f(x) = 0
x<1 ou x>6/5 ⇒ f(x)>0
1<x<6/5 ⇒ f(x)<0