What is the extreme f(x,y)=-x^4+2y^4+18x^2-36y^2+2 ?

The extreme f(x,y)=-x^4+2y^4+18x^2-36y^2+2 is

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extreme f(x,y)=-x^4+2y^4+18x^2-36y^2+2

extreme points

f (x, y) = - x^{4} + 2 y^{4} + 18 x^{2} - 36 y^{2} + 2

Saddle (0, 0), Minimum (0, - 3), Minimum (0, 3), Maximum (- 3, 0), Saddle (- 3, - 3), Saddle (- 3, 3), Maximum (3, 0), Saddle (3, - 3), Saddle (3, 3)

Solution steps

Find the critical points:

(0, 0), (0, - 3), (0, 3), (- 3, 0), (- 3, - 3), (- 3, 3), (3, 0), (3, - 3), (3, 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 y^{2} - 72

D (x, y) = (24 y^{2} - 72) (- 12 x^{2} + 36)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, - 3) :

Minimum

Check critical point

(0, 3) :

Minimum

Check critical point

(- 3, 0) :

Maximum

Check critical point

(- 3, - 3) :

Saddle

Check critical point

(- 3, 3) :

Saddle

Check critical point

(3, 0) :

Maximum

Check critical point

(3, - 3) :

Saddle

Check critical point

(3, 3) :

Saddle

Saddle (0, 0), Minimum (0, - 3), Minimum (0, 3), Maximum (- 3, 0), Saddle (- 3, - 3), Saddle (- 3, 3), Maximum (3, 0), Saddle (3, - 3), Saddle (3, 3)

f (x) = 7 x + \frac{1}{x}

x, y = - \frac{1}{2} f x_{+ 1}

5 cos (x)

f (x) = \frac{3}{5} (x^{2} - 1)^{\frac{2}{3}}

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

What is the extreme f(x,y)=-x^4+2y^4+18x^2-36y^2+2 ?
The extreme f(x,y)=-x^4+2y^4+18x^2-36y^2+2 is