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

The extreme f(x,y)=x^4-2x^2-y^2+3 is Maximum(0,0),Saddle(-1,0),Saddle(1,0)

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 2 (12 x^{2} - 4)

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Check critical point

(- 1, 0) :

Saddle

Check critical point

(1, 0) :

Saddle

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

ln (2 x^{2} - x)

f (x) = x^{2} - 3 x - 10

f (x) = x^{3} - x^{2} - 9 x + 9

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

f (x) = 6 x^{\frac{2}{3}} + 4 x

What is the extreme f(x,y)=x^4-2x^2-y^2+3 ?
The extreme f(x,y)=x^4-2x^2-y^2+3 is Maximum(0,0),Saddle(-1,0),Saddle(1,0)