What is the extreme f(x,y)=x^2+2xy^2+2y^2 ?

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

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extreme f(x,y)=x^2+2xy^2+2y^2

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 x + 4

D (x, y) = 2 (4 x + 4) - 16 y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(- 1, 1) :

Saddle

Check critical point

(- 1, - 1) :

Saddle

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

f (x) = \frac{x}{x}

f (x) = sec (x), - \frac{π}{6} \leq x \leq \frac{π}{3}

f (x) = \frac{3}{4} x^{4} + 15 x^{3} + 75 x^{2} - 23

f (x) = x^{3} - (\frac{3}{2}) x^{2}

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

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