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

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

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

extreme points

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

Saddle (- 1, 0), Minimum (- 1, - 2), Minimum (- 1, 2)

Solution steps

Find the critical points:

(- 1, 0), (- 1, - 2), (- 1, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2} - 16

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

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Saddle

Check critical point

(- 1, - 2) :

Minimum

Check critical point

(- 1, 2) :

Minimum

Saddle (- 1, 0), Minimum (- 1, - 2), Minimum (- 1, 2)

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

f (x) = x^{3} + 3 y^{3} + 3 x^{2} + 3 y^{2} + 24

f (x, y) = x + 4 y + \frac{2}{x y}

x e^{- x^{2}}

f (x) = 2 x^{3} + 12 x^{2} - 72 x + 11

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