What is the extreme f(x)=-x^4+2x^2y-1/2 y^2-4y+12 ?

The extreme f(x)=-x^4+2x^2y-1/2 y^2-4y+12 is Maximum(0,-4),Saddle(-2,4),Saddle(2,4)

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extreme f(x)=-x^4+2x^2y-1/2 y^2-4y+12

extreme points

f (x) = - x^{4} + 2 x^{2} y - \frac{1}{2} y^{2} - 4 y + 12

Maximum (0, - 4), Saddle (- 2, 4), Saddle (2, 4)

Solution steps

Find the critical points:

(0, - 4), (- 2, 4), (2, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 1

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

Check the sign of at each critical point

Check critical point

(0, - 4) :

Maximum

Check critical point

(- 2, 4) :

Saddle

Check critical point

(2, 4) :

Saddle

Maximum (0, - 4), Saddle (- 2, 4), Saddle (2, 4)

f (x y) = y^{2} ln (x^{2} y)

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

4 x^{3} - 12 x^{2} + 8 x

f (x) = 6 x^{2} - 6 x - 36

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

What is the extreme f(x)=-x^4+2x^2y-1/2 y^2-4y+12 ?
The extreme f(x)=-x^4+2x^2y-1/2 y^2-4y+12 is Maximum(0,-4),Saddle(-2,4),Saddle(2,4)