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

The extreme f(x,y)=y^3-2yx-xy^2+2x^2 is Saddle(0,0),Minimum(3/4 ,1),Saddle(2,2)

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

extreme points

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

Saddle (0, 0), Minimum (\frac{3}{4}, 1), Saddle (2, 2)

Solution steps

Find the critical points:

(0, 0), (\frac{3}{4}, 1), (2, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 2 x

D (x, y) = - 4 y^{2} + 16 y - 8 x - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{3}{4}, 1) :

Minimum

Check critical point

(2, 2) :

Saddle

Saddle (0, 0), Minimum (\frac{3}{4}, 1), Saddle (2, 2)

b - sin (x)

f (x) = x^{3} + 15 x - 2

f (x) = x^{4} + 2 x^{3} + 1 x^{2} + 1 x + 1

y = x^{2} - 3 x

y = 9 - 6 x^{2}, - 2 \leq x \leq 3

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