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k = 3√ 4 ξ ζ
p 3 √ 4 (y) = C(y − ξ)2 (y − ζ),
p ′ 3 √ 4 (y) = 2C(y − ξ)(y − ζ) + C(y − ξ)2 = C(y − ξ)(3y − 2ζ − ξ) = 3C(y − ξ)
y − 2
ξ
ζ − .
3 3
p ′ k (y) = 3(y2 − k) k = 3√ 4 p ′ k (y) = 3(y − 3√ 2)(y + 3√ 2)
C = 1 ξ = 3√ 2
ξ = − 3√ 2 p 3 √ 4 ( 3√ 2) = 0 p 3 √ 4 (− 3√ 2) = 0 ξ = 3√ 2
− 2
3 ζ − 3√ 2
3 = 3√ 2 ζ = −2 3√ 2 x = 3√ 4 (x, y) ∈ Γ
y ∈ {−2 3√ 2, 3√ 2}
x < 0 Γ y = ϕ1(x)
ϕ1(x) > 0 ∂yf(x, ϕ1(x)) > 0
ϕ1 ∈ C 1 (] − ∞, 0[; R)
Γ y = x x < 0
y > 0 (x, y) ∈ Γ
x > 0 y < 0 (x, y) ∈ Γ
ϕ2 :]0, +∞[→] − ∞, 0[ ϕ1
0 < x < 3√ 4 ϕ3, ϕ4 :]0, 3√ 4[→ ]0, +∞[ ϕ3(x) >
ϕ4(x) x
f(x, ϕi(x)) = 0 f
0 = lim sup
x→0 −
f(x, ϕ1(x)) = f(0, lim sup
x→0− ϕi(x)),
f(0, y) = 0 (0, y) ∈ Γ y = 0 lim sup ϕi(x))
lim inf
x→0− ϕi(x)) lim ϕi(x))
x→0−
x→0 −
lim
x→0− ϕ1(x) = lim
x→0 + ϕi(x) = 0, i = 2, 3, 4,
ϕ2(x) ≤ 0 x > 0 lim sup
x→ 3√ 4
0 = lim sup
x→ 3√ 4
ϕ2(x) ≤ 0
f(x, ϕ2(x)) = f( 3√ 4, lim sup
x→ 3√ ϕ2(x)),
4
f(0, y) = 0 (0, y) ∈ Γ y ∈ {−2 3√ 2, 3√ 2}
lim sup
x→ 3√ ϕ2(x) ∈ {−2
4
3√ 2, 3√ 2}.
lim sup
x→ 3√ ϕ2(x) = −2
4
3√ 2,
lim inf lim
x→ 3√ ϕ2(x) = −2
4
3√ 2
ϕ3, ϕ4 x
ϕi(x) ≤ 0 i = 3, 4 ϕ3 ϕ4
lim sup
x→ 3√ 4
( 3√ 4, 3√ 2)