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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 2 (6 x - 6 y) - 36 x^{2}

Check the sign of at each critical point

Check critical point

(0, - 1) :

Minimum

Check critical point

(- \frac{2}{3}, - \frac{1}{3}) :

Saddle

Check critical point

(1, \frac{1}{2}) :

Saddle

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

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

f (x) = {4 - 2 x, x \leq 1}

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

f (x) = - 300 x^{2} + 13500 x - 5400

f (x) = sin (x), 0 \leq x < \frac{π}{2}

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