bonjour,
si A(-10;110)∈f(x)
alors
yA=(xa)²-xa
110=(-10)²-10
110=100-10
faux
A∉f(x)
si B(4.2) ∈ f(x)
alors
yB=xB²-xb
2=4²-4
2=16-4
faux
B ∉(fx)
si A(10;110) ∈g(x)
alors
yA=-xA²+3(xA)+6
110=-(10²)+3(10)+6
110=-100+30+6
faux
A ∈ g(x)
si B (4.2)∈ g(x)
alors
yB=-(xB²)+3(xB)+6
2=-4²+3(4)+6
2=-16+12+6
2=-16+18
exact
B ∈ g(x)
f(x)=g(x)
x=-1 y= 2
x=3 y=6
f(x)=g(x)
x²-x=-x²+3x+6
x²+x²-x-3x-6=0
2x²-4x-6=0
Δ=4²-4(2)(-6)
Δ=16-4(-12)
Δ=16+48
Δ=64
√Δ=8
x=(4+8)/4 x=12/4 x=3
x=(4-8)/4 x=-4/4 x=-1
f(x)≥g(x) x ∈ ]-∞:-1] ∪ [3;+∞[
f(x)<g(x) x ∈]-1:3[
f(x)≤g(x) x ∈ [-1:3]
f(x)<6 x ∈ ]-2:3[
g(x)≤f(x) x ∈]-∞:-1] ∪ [3,+∞[
g(x)≤f(x)<6 x ∈ ( ]-2:3[ ) ∩ ( ]-∞;-1] ∪[3;+∞[
g(x)≤f(x)<6 x ∈ ]-2;-1]
