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α ∈ R (0, 0)
f(x, y) = 2 + αx 2 + 4xy + (α − 3)y 2 + (2x + y) 4 .
f ∈ C 2 f(0, 0) = 2 f
∂xf(x, y) = 2αx + 4y + 8(2x + y) 3
∂yf(x, y) = 4x + 2(α − 3)y + 4(2x + y) 3
∂xxf(x, y) = 2α + 48(2x + y) 2
∂xyf(x, y) = 4 + 24(2x + y) 2
(0, 0)
D 2
2α 4
f(0, 0) =
4 2(α − 3)
∂yyf(x, y) = 2(α − 3) + 12(2x + y) 2 .
, detD 2 f(0, 0) = (α − 4)(α + 1).
λ1, λ2 ∈ R −1 <
α < 4 detD 2 f(0, 0) < 0
detD 2 f(0, 0) > 0 1, 1
α > 4 detD 2 f(0, 0)
α < −1
λ 2 − 2(2α − 3)λ + 4(α 2 − 3α − 4) = 0
λ = 2α − 3 ± 4α 2 + 9 − 12α − 4α 2 + 12α + 16 = 2α − 3 ± 5,
λ1 = 2(α + 1) λ2 = 2(α − 4) λ1 > λ2 λ2 > 0 α > 4
(0, 0) λ1 < 0 α < −1
(0, 0) −1 < α < 4 λ2 < 0 λ1 > 0
α ∈ {−1, 4}
α = 4 (x, y) = (0, 0)
f(x, y) = 2 + 4x 2 + 4xy + y 2 + (2x + y) 4 = 2 + (2x + y) 2 + (2x + y) 4 > 2,
f(x, −2x) = 2
x
α = −1
f(x, y) = 2 − x 2 + 4xy − 4y 2 + (2x + y) 4 = 2 − (x − 2y) 2 + (2x + y) 4 .
γ1(t) = (2t, t) t > 0 f ◦ γ1(t) = 2 + 5 4 t 4 > 2
γ2(t) = (t, −2t) t > 0 f ◦ γ2(t) = 2 − 25t 2 < 2 t → 0
γ1(t) → (0, 0) γ2(t) → (0, 0) (0, 0) γ1(t) f
f(0, 0) γ2(t) f f(0, 0)
(0, 0)