What is the extreme f(x,y)=-x^4+4(x^2-y^2)-3 ?

The extreme f(x,y)=-x^4+4(x^2-y^2)-3 is Saddle(0,0),Maximum(-sqrt(2),0),Maximum(sqrt(2),0)

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

extreme points

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

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}} = - 8

D (x, y) = - 8 (- 12 x^{2} + 8)

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)

f (x, y) = \frac{- x - ln ( x ^{2} + y ^{2} - 81 )}{666}

f (x) = sin (2 x) - x, - \frac{π}{2} \leq x \leq \frac{π}{2}

f (x) = 5 x^{4} - x^{5} + 8

f (x) = 5 x^{4} - x^{5} + 7

f (x) = 5 x^{4} - x^{5} + 4

What is the extreme f(x,y)=-x^4+4(x^2-y^2)-3 ?
The extreme f(x,y)=-x^4+4(x^2-y^2)-3 is Saddle(0,0),Maximum(-sqrt(2),0),Maximum(sqrt(2),0)