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SaSa98
@SaSa98
June 2021
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Merci énormément pour votre aide!!!
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slyz007
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1) f(x)≤0
5-x²≤0
⇔x²≥5
⇔x≤-√5 ou x≥√5
S=]-oo;-√5]U[√5;+oo[
2) On sait que g(x) est de la forme g(x)=ax+b
Or g(-1)=2 donc 2=-a+b soit b=a+2
g(2)=5 donc 5=2a+b=2a+a+2 soit 3a=3 et a=1
Donc b=1+2=3
g(x)=x+3
3a) Graphiquement, on voit que f(x)≥g(x) si x∈[-2;1]
3b) 5-x²≥x+3
⇔x²+x-2≤0
Δ=1²+4*1*2=9
√Δ=3
x1=(-1-3)/2=-2 et x2=(-1+3)/2=1
Donc x²+x-2=(x+2)(x-1)
Donc (x+2)(x-1)≤0 si x∈[-2;1]
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Verified answer
1) f(x)≤05-x²≤0
⇔x²≥5
⇔x≤-√5 ou x≥√5
S=]-oo;-√5]U[√5;+oo[
2) On sait que g(x) est de la forme g(x)=ax+b
Or g(-1)=2 donc 2=-a+b soit b=a+2
g(2)=5 donc 5=2a+b=2a+a+2 soit 3a=3 et a=1
Donc b=1+2=3
g(x)=x+3
3a) Graphiquement, on voit que f(x)≥g(x) si x∈[-2;1]
3b) 5-x²≥x+3
⇔x²+x-2≤0
Δ=1²+4*1*2=9
√Δ=3
x1=(-1-3)/2=-2 et x2=(-1+3)/2=1
Donc x²+x-2=(x+2)(x-1)
Donc (x+2)(x-1)≤0 si x∈[-2;1]