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Cyasminechabine
@Cyasminechabine
April 2019
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.slvp aider moi c'est a rendre !!
Dérivée
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Ex 1 :
1) f(x)=-3/x+3x²+7
f'(x)=3/x²+6x
2) h(x)=(x²+3x-7)/(x+5)
h'(x)=((2x+3)(x+5)-(x²+3x-7))/(x+5)²
=(2x²+13x+15-x²-3x+7)/(x+5)²
=(x²+10x+22)/(x+5)²
3) g(x)=-π/x
g'(x)=π/x²
4) k(x)=(8-x)(2x²-x+7)
k'(x)=-(2x²-x+7)+(8-x)(4x-1)
=-2x²+x-7-4x²+32x-8+x
=-6x²+34x-15
ex 2:
f(x)=2/3x³-3/2x²-2x+1
f'(x)=2x²-3x-2=(2x+1)(x-2)
donc f est croissante sur ]-∞;-1/2] et sur [2;+∞[
f est décroissante sur [-1/2;2]
f(x)≥0 donne 4x³-9x²-12x+6≥0
soit S=[-1,22;0,4] U [3,07;+∞[
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Verified answer
Ex 1 :1) f(x)=-3/x+3x²+7
f'(x)=3/x²+6x
2) h(x)=(x²+3x-7)/(x+5)
h'(x)=((2x+3)(x+5)-(x²+3x-7))/(x+5)²
=(2x²+13x+15-x²-3x+7)/(x+5)²
=(x²+10x+22)/(x+5)²
3) g(x)=-π/x
g'(x)=π/x²
4) k(x)=(8-x)(2x²-x+7)
k'(x)=-(2x²-x+7)+(8-x)(4x-1)
=-2x²+x-7-4x²+32x-8+x
=-6x²+34x-15
ex 2:
f(x)=2/3x³-3/2x²-2x+1
f'(x)=2x²-3x-2=(2x+1)(x-2)
donc f est croissante sur ]-∞;-1/2] et sur [2;+∞[
f est décroissante sur [-1/2;2]
f(x)≥0 donne 4x³-9x²-12x+6≥0
soit S=[-1,22;0,4] U [3,07;+∞[