Determine all the local minima, local maxima, and saddles of the function?

Question

Determine all the local minima, local maxima, and saddles of the function f(x, y)=x^4-x^2-2xy+y^2.

The gradient for f=(4x^3-2x-2y, -2x+2y)=(0, 0)

x=y 4x^3-4x=0 x=0, -1, 1

Critical points: (0, 0) and (1, 1) and (-1, -1)

By the Hessian matrix,

12x^2-2    -2

-2              2

At (0, 0),

-2,   -2

-2,   2

The determinant is -8

At (1, 1) and (-1, -1)

10    -2

-2      2

=16

But how do I know which is the saddle, minima and maxima?

Answer 1

If the 1^st and 2^nd order partial derivatives of a function ƒ are continuous throughout an open region containing (a, b) and ƒ_x=ƒ_y=0, the critical point (a, b) may be classified using the 2^nd derivative test:

(i) if ƒ_xx<0 and ƒ_xxƒ_yy - ƒ²_xy>0 at (a, b), then (a, b) is a local maximum

(ii) if ƒ_xx>0 and ƒ_xxƒ_yy - ƒ²_xy>0 at (a, b), then (a, b) is a local minimum

(iii) if ƒ_xxƒ_yy - ƒ²_xy<0 at (a, b), then (a, b) is a saddle point

(iv) if ƒ_xxƒ_yy - ƒ²_xy=0, then the test is inconclusive

   1^st partials     ƒ_x = 4x³ - 2x - 2y   ,     ƒ_y = -2x + 2y

    Critical points:     (0, 0) ,  (-1, -1) ,  (1, 1)

    2^nd partials:     ƒ_xx = 12x² - 2   ,   ƒ_yy = 2   ,   ƒ_xy = -2

   at (0, 0):   ƒ_xx = -2   and   ƒ_xxƒ_yy - ƒ²_xy = (-2)(2) - (-2)² = -4 - 4 = -8

          since ƒ_xxƒ_yy - ƒ²_xy<0, then (0, 0) is a saddle point

   at (-1, -1):   ƒ_xx = 10   and   ƒ_xxƒ_yy - ƒ²_xy = (10)(2) - (-2)² = 20 - 4 = 16

          since ƒ_xxƒ_yy - ƒ²_xy>0 and ƒ_xx>0, then (-1, -1) is a local minimum

   at (1, 1):   ƒ_xx = 10   and   ƒ_xxƒ_yy - ƒ²_xy = (10)(2) - (-2)² = 20 - 4 = 16

          since ƒ_xxƒ_yy - ƒ²_xy>0 and ƒ_xx>0, then (1, 1) is a local minimum