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

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

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

extreme points

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

Saddle (0, 0), Maximum (- 1, - 1), Maximum (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) = 192 y^{2} - 64

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Check critical point

(1, 1) :

Maximum

Saddle (0, 0), Maximum (- 1, - 1), Maximum (1, 1)

f (x) = \frac{2 ( - x + 1 )}{x ^{3}}

x^{4} + 2 x^{3} + 6 x^{2} - 9 x + 12

f (x) = ∣ 2 x - 4 ∣

y (t, x) = - 2 t + x

f (x) = 2 x^{2} + 3 x - 5, - 4 \leq x \leq 4

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