What is the extreme f(x,y)=4y^3+x^2-12y^2-36y+2 ?

The extreme f(x,y)=4y^3+x^2-12y^2-36y+2 is Minimum(0,3),Saddle(0,-1)

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extreme f(x,y)=4y^3+x^2-12y^2-36y+2

extreme points

f (x, y) = 4 y^{3} + x^{2} - 12 y^{2} - 36 y + 2

Minimum (0, 3), Saddle (0, - 1)

Solution steps

Find the critical points:

(0, 3), (0, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 y - 24

D (x, y) = 2 (24 y - 24)

Check the sign of at each critical point

Check critical point

(0, 3) :

Minimum

Check critical point

(0, - 1) :

Saddle

Minimum (0, 3), Saddle (0, - 1)

f (0.7) = 17 e^{- \frac{1}{6} x y}

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

f (x, y) = - 11 y^{2} + (x + 16)^{2} + 1

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

f (x) = 4 x^{3} - 6 x^{2} - 72 x + 30

What is the extreme f(x,y)=4y^3+x^2-12y^2-36y+2 ?
The extreme f(x,y)=4y^3+x^2-12y^2-36y+2 is Minimum(0,3),Saddle(0,-1)