(∀x∈]0,1[∪]1,+∞[),x2−x
2(x)≤x2
x
dt
(t)≤x2−x
(x)
]0,1[∪]1,+∞[ φ
φ(1) = 1 φ(0) = 0 φ(x) = x−1
(x)
(∀x∈]0,1[∪]1,+∞[), x −1−x(x)<0
[0,+∞[ φ
φ
(∀x∈]0,1[∪]1,+∞[), x −1≤x2
x
φ(t)
(t)dt ≤x2−1
2
]0,1[∪]1,+∞[ φ
f(0) = f(1) = 0 f(x) = −(x+ 1) + x2
x
dt
(t)
f
f
(∀x∈]0,1[∪]1,+∞[),x2
x
φ(t)
(t)dt =f(x) + x+ 1
2
x0= 1 f
f′(α) = 0 ]0,1[ α
[0,+∞[ f
]0,1[∪]1,+∞[ ψ
ψ(0) = ψ(1) = 1 ψ(x) = (x+ 1) −x2
x
dt
(t)
(∀x∈[0,+∞[) , ψ(x) = (−f(x))
[0,+∞[ ψ
O,⃗
i,⃗
j ψ
ψ(α)≃1,2α≃0,3
β ψ ∴
β∈3
2,2
(∀n∈N∗) (∃!γn>0) ,ψ(γn)
γn
=n
n+ 1 ·ψ(α)
α
(γn)n≥1
(Fx) : t3+xt + 1 = 0 [0,+∞[ x
f(x) R (Fx)
[0,+∞[ f
]0,+∞[ f
]0,+∞[ f
.(∀x > 0) , f′(x) + f(x)
x+ 3f2(x)= 0
x→+∞f(x) f
f(1) = 1 f(t) = t−1
(t) [1,+∞[
F(1) = −(2) F(x) = x
x2
f(t)
(t)dt
[1,+∞[ f
.(∀x∈]1,+∞), F (x)−F(1) = x(x−1)(x+ 2)
2·f(x)+x
x22 + 1
tf(t)dt
1 F
(∀x∈]1,+∞[) (∃cx∈x, x2), F (x) = x−x2·cx−1
2(cx)
.∀x≥√2,+∞,F(x)
x+f(x)
22
≤0
x→+∞
F(x)
x
x→+∞F(x)
F ]1,+∞[ x F′(x)
(∀x∈]1,+∞) (∃(c, d)∈]1, x[) (c > d),F(x) + (2)
x−1=−c+ 2
2·d2
x0= 1 F
O,⃗
i,⃗
j F
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
1
/
33
100%