pdf (it, 2.898,97 KB, 25/01/13)

H
A 2 × 2
λ 2 − tr(A) λ + det A = 0,
tr(A) A A
f : Ω → R Ω R 2 (x0, y0) ∈ Ω
∇f(x0, y0) = 0
· ∂ 2 xxf(x0, y0) > 0 det D 2 (x0, y0) > 0 (x0, y0)
· ∂ 2 xxf(x0, y0) < 0 det D 2 (x0, y0) > 0 (x0, y0)
· det D 2 (x0, y0) < 0 (x0, y0)
· det D 2 (x0, y0) = 0
R
f(x) = |x| 0
f(x, y) = 2x 3 + y 3 − 3x 2 − 3y
f(x, y) = x 3 + y 3 − (1 + x + y) 3
f(x, y) = cos x sin y
f(x, y) = x 4 + x 2 y + y 2 + 3
f(x, y) = x 4 + y 4 − 2(x 2 + y 4 ) + 4xy
f
∂xf(x, y) = 6x 2 − 6x = 0 =⇒ x ∈ {0, 1},
∂yf(x, y) = 3y 2 − 3 = 0 =⇒ y ∈ {1, −1}.
(0, ±1) (1, ±1)
f
D 2
∂2 f(x, y) =: xxf(p) ∂2 xyf(p)
∂2 yxf(p) ∂2
,
yyf(p)
∂ 2 xxf(x, y) = 12x − 6, ∂ 2 yxf(x, y) = ∂ 2 xyf(x, y) = 0, ∂ 2 yyf(x, y) = 6y.
D 2 f(x, y) =:
12x − 6 0
0 6y
D 2 f(x, y) λ1(x, y) = 12x − 6 λ2(x, y) = 6y
λ1(0, 1) = −6 λ2(0, 1) = 6 (0, 1)
λ1(0, −1) = −6 λ2(0, −1) = −6 (0, −1)
λ1(1, 1) = 6 λ2(1, 1) = 6 (1, 1)
λ1(1, −1) = 18 λ2(1, −1) = −6 (1, −1)
.