Compute the discriminant D(x,y)D(x,y) of the function. f(x,y)=x3+y4−6x−2y2+3 f(x,y)=x3+y4−6x−2y2+3 (Express numbers in exact form. Use symbolic notation and fractions where needed.) D(x,y)=D(x,y)= Which of these points are saddle points? (2⎯⎯√,0)(2,0) (2⎯⎯√,1)(2,1) (−2⎯⎯√,1)(−2,1) (2⎯⎯√,−1)(2,−1) (−2⎯⎯√,−1)(−2,−1) (−2⎯⎯√,0)(−2,0) Which of these points are local minima?

The discriminant of function [tex]f(x,y)=x^3 + y^4 - 6x-2y^2+3[/tex] is 72xy³ – 24x and points (√2,0), (-√2, 1) and (-√2, -1) are saddle point, and points (√2, 1) and (√2, -1) are local minima.

A function is an expression in terms of one or more variable.

If f(x, y)  is a two-dimensional function that has a local extremum at a point ([tex]x_o, y_o[/tex])  and has continuous partial derivatives at this point, then [tex]f_x (x_o, y_o)=0[/tex] and [tex]f_y (x_o, y_o)=0[/tex]. The second partial derivatives test classifies the point as a local maximum or local minimum.

Then

1. If D > 0 and [tex]f_{xx}(x_o, y_o) > 0[/tex] , the point is a local minimum.

2. If D > 0 and [tex]f_x_x (x_o, y_o) < 0[/tex], the point is a local maximum.

3. If D < 0, the point is a saddle point.

4. If D = 0, higher order tests must be used.

Given that

[tex]f(x,y)=x^3 + y^4 - 6x-2y^2+3[/tex]

[tex]f_x = 3x^2-6[/tex]

[tex]f_x = 0[/tex] implies that [tex]x = \pm\sqrt2[/tex]

[tex]f_y= 4y^3-4y[/tex]

[tex]f_y= 0[/tex]  implies that y = 0; y = [tex]\pm[/tex]1.

Thus, the critical points are (√2,0), (√2,1), (√2, -1);(-√2,0), (-√2,1); (√2, -1).

[tex]f_{xx} = 6x[/tex]; [tex]f_y_y = 12y^3 - 4[/tex], [tex]f_{xy}= 0[/tex]

D(x, y) = [tex]f_{xx}f_{yy} - f_{xy}^2[/tex]

          = 6x × (12y³ – 4)

D(√2, 0) = -24√2 < 0  implies (√2,0)  is a saddle point.

D(√2, 1) = 48√2 > 0 ; [tex]f_{xx}(2, 1) = 6\sqrt2 > 0[/tex] implies (√2, 1)  is local minimum.

D(√2, -1) = 48√2 > 0; [tex]f_{xx}(2, -1) = 6\sqrt2 > 0[/tex]  implies (√2, -1)  is local minimum.

D(-√2, 0) = 24√2 > 0; [tex]f_{xx}(-2,0) = -6\sqrt2 < 0[/tex]  implies (-√2, 0)  is local maximum.

D(-√2, 1) = -48√2 < 0  implies (-√2, 1)  is a saddle point.

D(-√2, -1) = -48√2 < 0  implies (-√2, -1)  is a saddle point.

Thus, (√2,0), (-√2, 1) and (-√2, -1) are saddle point, and (√2, 1) and (√2, -1) are local minima.

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