On calcule maintenant les dérivées partielles secondes : \[\frac{\partial^2 f}{\partial x^2}(x,y) = -2y,\quad \frac{\partial^2 f}{\partial y^2}(x,y) = 1,\quad \frac{\partial^2 f}{\partial x \partial y}(x,y) = \frac{\partial^2 f}{\partial y \partial x}(x,y) = -2x.\]

Le déterminant de la matrice hessienne est : \[\Delta_f(x,y) = \begin{vmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{vmatrix} = -2y-4x^2.\]

En \((0,-1)\) : \[\frac{\partial^2 f}{\partial x^2}(0,-1) = 2,\quad \frac{\partial^2 f}{\partial x \partial y}(0,-1) = 0,\quad \frac{\partial^2 f}{\partial y^2}(0,-1) = 1.\] Donc \[\Delta_f(0,-1) = 2 \cdot 1 - 0 = 2 > 0,\quad \text{et} \quad \frac{\partial^2 f}{\partial x^2}(0,-1) > 0\] \[\Rightarrow \text{la fonction}\ f\ \text{admet un minimum local en} (0,-1).\] En \((1,0)\) : \[\frac{\partial^2 f}{\partial x^2}(1,0) = 0,\quad \frac{\partial^2 f}{\partial x \partial y}(1,0) = -2,\quad \frac{\partial^2 f}{\partial y^2}(1,0) = 1.\] Donc \[\Delta_f(1,0) = 0 \cdot 1 - (-2)^2 = -4 < 0 \Rightarrow \text{la fonction} f \text{ admet un point selle en}\ (1,0).\]

En \((-1,0)\) : \[\frac{\partial^2 f}{\partial x^2}(-1,0) = 0,\quad \frac{\partial^2 f}{\partial x \partial y}(-1,0) = 2,\quad \frac{\partial^2 f}{\partial y^2}(-1,0) = 1.\] Donc \[\Delta_f(-1,0) = 0 \cdot 1 - (2)^2 = -4 < 0 \Rightarrow \text{la fonction} f \text{ admet un point selle en}\ (-1,0).\]

Recherche des extremums globaux