What is the extreme f(x,y)=x^3+y^2-3x-4y ?

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

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extreme f(x,y)=x^3+y^2-3x-4y

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 12 x

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Saddle

Check critical point

(1, 2) :

Minimum

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

\frac{x ^{2} - 11}{x + 6}

f (x, y) = x^{3} - 12 x - 3 y + y^{3}

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

f (x) = \frac{x ^{2} - 8}{x + 3}

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

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