What is the extreme f(x,y)=x^3-y^2-3x+12y ?

The extreme f(x,y)=x^3-y^2-3x+12y is Maximum(-1,6),Saddle(1,6)

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extreme f(x,y)=x^3-y^2-3x+12y

extreme points

f (x, y) = x^{3} - y^{2} - 3 x + 12 y

Maximum (- 1, 6), Saddle (1, 6)

Solution steps

Find the critical points:

(- 1, 6), (1, 6)

\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, 6) :

Maximum

Check critical point

(1, 6) :

Saddle

Maximum (- 1, 6), Saddle (1, 6)

f (x) = (x^{2} - 12) \cdot e^{- 2 x}

f (x) = 49 x^{2} - 10 x + 17

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

f (x, y) = - 3 y^{3} - 3 x y + 13 y - 2 \cdot 2 + 5 x + 19

f (x) = x y + y - 16 x

What is the extreme f(x,y)=x^3-y^2-3x+12y ?
The extreme f(x,y)=x^3-y^2-3x+12y is Maximum(-1,6),Saddle(1,6)