Esercizi svolti di analisi reale e complessa - Dipartimento di ...

f
∂xf(x, y) = 2x − y 2 = 0
∂yf(x, y) = −2xy = 0
⇐⇒
x = 0
y = 0 .
f(0, 0) = 0
Bδ(0, 0) 0 < δ < 1
P1 =
δ 2 , δ
2
e P2 = − δ
δ
,
2 2
f(P1) = − 3
4 δ2 < 0 e f(P2) = δ2 δ3
+ > 0 .
4 8
K x = ± 1
2
x 2 + y 2 = 1
A =
− 1
2 ,
√
3
, B =
2
1
2 ,
√
3
, C = −
2
1
√ √
3 1 3
, − , D = , − .
2 2
2 2
f(A) = f(C) = 5
8
f(B) = f(D) = − 1
8 .
• L+
L± =
x = ± 1
√
3
, −
2 2
f+(y) = f(x, y) |L+
f ′ +(y) = −y
√
3
< y < .
2
1 1
= f(1 , y) = −
2 4 2 y2
E = ( 1
1
2 , 0) f(E) =
• L−
f−(y) = f(x, y) |L−
f ′ −(y) = y
1 1
= f(−1 , y) = +
2 4 2 y2
G = (− 1
1
2 , 0) f(G) =
;
4
4