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f(x, y) = f(y, x)
f
∂xf(x, y) = 3x 2 − 3(1 + x + y) 2 = 0
∂yf(x, y) = 3y 2 − 3(1 + x + y) 2 = 0.
x = ±y x = −y 3y 2 − 3 = 0 y = ±1
(−1, 1) (1, −1) x = y 3y 2 − 3(1 + 4y 2 + 2y) = 0 y = −1
y = −1/3 (−1, −1) (−1/3, −1/3)
(x, y) ↦→ (y, x)
∂ 2 xxf(x, y) = 6x − 6(1 + x + y), ∂ 2 yyf(x, y) = 6y − 6(1 + x + y), ∂ 2 xyf(x, y) = ∂ 2 yxf(x, y) = −6(1 + x + y).
D 2
−12 −6
f(−1, 1) =:
, D
−6 0
2
0 −6
f(1, −1) =:
,
−6 −12
D 2
0 6
f(−1, −1) =: , D
6 0
2
−4 −2
f(−1/3, −1/3) =:
−2 −4
(−1, 1) λ 2 + 12λ − 36 = 0 λ1 =
−6 + 6 √ 2 λ1 = −6 − 6 √ 2
(1, −1) (−1, 1)
(−1, −1) λ 2 − 36 = 0 λ = ±6
(−1/3, −1/3) λ 2 + 8λ + 12 = 0
λ1 = −6 λ2 = −2
2π
[0, 2π[×[0, 2π[
∂xf(x, y) = − sin x sin y ∂yf(x, y) = cos x cos y
∂xf(x, y) = 0 x ∈ {0, π} y ∈ {0, π}
∂yf(x, y) = 0 x ∈ {π/2, 3π/2} y ∈ {π/2, 3π/2}
(0, π/2) (0, 3π/2) (π, π/2) (π, 3π/2) (π/2, 0) (3π/2, 0) (π/2, π) (3π/2, π)
∂ 2 xxf(x, y) = − cos x sin y, ∂ 2 yyf(x, y) = − cos x sin y, ∂ 2 xyf(x, y) = − sin x cos y.
D 2
−1 0
f(0, π/2) =:
, D
0 −1
2
1 0
f(0, 3π/2) =: ,
0 1
D 2
1 0
f(π, π/2) =: , D
0 1
2
−1 0
f(π, 3π/2) =:
,
0 −1
D 2
0 −1
f(π/2, 0) =:
, D
−1 0
2
0 1
f(3π/2, 0) =: ,
1 0
D 2
0 1
f(π/2, π) =: , D
1 0
2
0 −1
f(3π/2, π) =:
.
−1 0