1666724786535_math analyse

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2.
= 2y. e x
f ′ x (x 0; y 0
) = 0
f ′ y (x 0; y 0
) = 0
⇨
e x (2x + x 2 + y 2 ) = 0
2y. e x = 0
(2x + x 2 + y 2 ) = 0
(2x + x 2 ) = 0
⇨
⇨
2y = 0
y = 0
(2x + x 2 ) = 0
⇨ x(2 + x) = 0
x = 0 ou x = −2
Alors les points critiques de f sont (0,0)et (−2,0)
3. f ′′ x 2 (x, y) = [e x . (2x + x 2 + y 2 )] ′
= (e x )′. (2x + x 2 + y 2 ) + (2x + x 2 + y 2 )′. e x
f ′′ y 2 (x, y) = [2y. e x ] ′ = 2.1. e x = 2e x
= e x . (2x + x 2 + y 2 ) + e x . (2 + 2x)
= e x (x 2 + y 2 + 4x + 2)
f ′′ yx (x, y) = f′′ xy (x, y) = ex (2x + x 2 + y 2 ) = e x . 2y = 2ye x
e x (x 2 + y 2 + 4x + 2)
2ye x
H =
2ye x
2e x