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lulul64
@lulul64
January 2021
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Bonjour pouvez vous m'aider pour ces exercices svp c'est urgent
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aur70
Verified answer
Bonjour
exo2
1/ a/ -2(x-3)²+2=-2(x²-6x+9)+2=-2x²+12x-18+2=2x²+12x-16=f(x)
b/ f'(x)=-4x+12=-4(x-3)
x≤3 ⇔ x-3≤0 ⇔4(x-3)≤0 ⇔ -4(x-3)≥0 si x≤3 donc f est croissante
c/ x ║ -∞ ║ 3 ║ +∞
f'(x) ║ + ║ 0 ║ -
f(x) ║ croiss ║ décr
2/ a/ (-x+4)(2x-4)=-2x²+4x+8x-16=-2x²+12x-16=f(x)
b/ f(x)=0 ⇔ x=4 ou x=2
x ║ -∞ 2 4 +∞
(-x+4) ║ + + 0 -
(2x-4) ║ - 0 + +
f(x) ║ - + -
exo3
a/ f(1/2)=(2*1/2 +1)/(1/2 -1)=2/(-1/2)=-4
f(-3/2)=(-2*3/2 +1)/(-3/2 -1)=(-2)/(-5/2)=4/5
b/ f(x)=34 ⇔ (2x+1)/(x-1)=34 ⇔ 2x+1=34x-34 ⇔32x=35 ⇔x=35/32
c/ 2+3/(x-1)=[2(x-1)+3]/(x-1)=(2x-2+3)/(x-1)=(2x+1)/(x-1)=f(x)
d/ f'(x)=-1/(x-1)²<0 donc f est strictement décroissante
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lulul64
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HELP ME PLEASE !!!!! URGENTISSIME !!!!
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Lista de comentários
Verified answer
Bonjourexo2
1/ a/ -2(x-3)²+2=-2(x²-6x+9)+2=-2x²+12x-18+2=2x²+12x-16=f(x)
b/ f'(x)=-4x+12=-4(x-3)
x≤3 ⇔ x-3≤0 ⇔4(x-3)≤0 ⇔ -4(x-3)≥0 si x≤3 donc f est croissante
c/ x ║ -∞ ║ 3 ║ +∞
f'(x) ║ + ║ 0 ║ -
f(x) ║ croiss ║ décr
2/ a/ (-x+4)(2x-4)=-2x²+4x+8x-16=-2x²+12x-16=f(x)
b/ f(x)=0 ⇔ x=4 ou x=2
x ║ -∞ 2 4 +∞
(-x+4) ║ + + 0 -
(2x-4) ║ - 0 + +
f(x) ║ - + -
exo3
a/ f(1/2)=(2*1/2 +1)/(1/2 -1)=2/(-1/2)=-4
f(-3/2)=(-2*3/2 +1)/(-3/2 -1)=(-2)/(-5/2)=4/5
b/ f(x)=34 ⇔ (2x+1)/(x-1)=34 ⇔ 2x+1=34x-34 ⇔32x=35 ⇔x=35/32
c/ 2+3/(x-1)=[2(x-1)+3]/(x-1)=(2x-2+3)/(x-1)=(2x+1)/(x-1)=f(x)
d/ f'(x)=-1/(x-1)²<0 donc f est strictement décroissante