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x = k |k| > 4
x = 4 (4, 0)
f
f = 0
v = ρ 2 = x 2 + y 2 0 ≤ v ≤ 16
px(v) = 4(v − x) 3 − 27v 2 = 0,
v v ≥ x 2
x x = v − 3v 2/3 / 3√ 4 v ≥ x 2
h(v) = v − 3v 2/3 / 3√ 4 x = h(v) 0 ≤ v ≤ 16
v ≥ h 2 (v)
h ′ (v) = 1 − 2 1/3 /v 1/3 v = 2
h
0 < v < 2 2 < v < 16
x = h(v) 0 ≤ v ≤ 16 v ≥ x 2 − √ v ≤ x ≤ √ v
h(v) = 0 v = 0 v = 27/4 0 < 2 < 27/4 < 16 h
]0, 27/4[ ]27/4, 16[
v h(v) h(16) = 4
− √ 16 ≤ 4 ≤ √ 16 x
f(x, y) = 0 4 f(4, y) = 0
y = v − h 2 (v) v = 16 y = 0
x = h(v) f(x, y)
h v = 2 h(2) = −1 − √ 2 < −1
− √ v v > 2 h(v)
2 < v < 16 − √ v ≤ h(v) = x
(x, y) ∈ Γ x ≥ −1 −1
x Γ
f(−1, y) = 0 5 − 6y 2 − 3y 4 + 4y 6 = 0 t = y 2 5 − 6t −
3t 2 + 4t 3 = 0 5 ±1, ±5 1
5 − 6t − 3t 2 + 4t 3 x − 1 4t 2 + t − 5
1 −5/4
f(−1, y) = 0 t = y 2 = 1 y = ±1 ∂xf(1, ±1) = 0
x y
h(v) − √ v 2 < v < 16
2
4 2 < v < 16 h(v)
0 < x < 4 v h(v) = x x0 ≥ 0
f(x0, y) = 0 y
y = ± 1 − h 2 (v)
h 2
[0, 2] [2, 16] −1 < x < 0 v
y f(x, y) = 0
y 2 Γ y 2 = v − h 2 (v) 1 −
2h(v)h ′ (v) = 5 3√ 2v 2/3 −2v−3·2 2/3 3√ v+1 y 2
x = −1/2 0 v = 1/4 1 − 2h(v)h ′ (v)
v0 = 1/4
p(v) = 5 3√ 2v 2/3 − 2v − 3 · 2 2/3 3√ v + 1 p(1/4) = 0 v = 2t 3
p(t) = −4t 3 + 10t 2 − 6t + 1 1/4 = 2t 3
t = 1/2 p(t) t − 1/2