What is the extreme f(x,y)=3x^3+y^2-9x+4y ?

The extreme f(x,y)=3x^3+y^2-9x+4y is Saddle(-1,-2),Minimum(1,-2)

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extreme f(x,y)=3x^3+y^2-9x+4y

extreme points

f (x, y) = 3 x^{3} + y^{2} - 9 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) = 36 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)

f (x) = \frac{- 2 x ^{2} + 14 x - 24}{x ^{2} + 2 x}

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

f (x) = 1 - x - x^{2}

f (x) = 32 y^{2} + x^{2} - x^{2} y

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

What is the extreme f(x,y)=3x^3+y^2-9x+4y ?
The extreme f(x,y)=3x^3+y^2-9x+4y is Saddle(-1,-2),Minimum(1,-2)