What is the extreme f(x,y)=-x^4+8x^2-y^2-1 ?

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

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extreme f(x,y)=-x^4+8x^2-y^2-1

extreme points

f (x, y) = - x^{4} + 8 x^{2} - y^{2} - 1

Saddle (0, 0), Maximum (- 2, 0), Maximum (2, 0)

Solution steps

Find the critical points:

(0, 0), (- 2, 0), (2, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2

D (x, y) = - 2 (- 12 x^{2} + 16)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, 0) :

Maximum

Check critical point

(2, 0) :

Maximum

Saddle (0, 0), Maximum (- 2, 0), Maximum (2, 0)

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