What is the extreme f(x,y)=-8xy+2x^4+2y^4 ?

The extreme f(x,y)=-8xy+2x^4+2y^4 is Saddle(0,0),Minimum(-1,-1),Minimum(1,1)

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extreme f(x,y)=-8xy+2x^4+2y^4

extreme points

f (x, y) = - 8 x y + 2 x^{4} + 2 y^{4}

Saddle (0, 0), Minimum (- 1, - 1), Minimum (1, 1)

Solution steps

Find the critical points:

(0, 0), (- 1, - 1), (1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 y^{2}

D (x, y) = 576 x^{2} y^{2} - 64

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, - 1) :

Minimum

Check critical point

(1, 1) :

Minimum

Saddle (0, 0), Minimum (- 1, - 1), Minimum (1, 1)

f (x) = 2 x - sin (x), 0 \leq x \leq π

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 8

13 + 4 x - x^{2}

f (x) = \frac{ln ( x )}{x ^{3}}

f (x, y) = 2 x^{2} - 3 x y + 4

What is the extreme f(x,y)=-8xy+2x^4+2y^4 ?
The extreme f(x,y)=-8xy+2x^4+2y^4 is Saddle(0,0),Minimum(-1,-1),Minimum(1,1)