What is the extreme f(x,y)=3x^3-12x+2y^2-8y+7 ?

The extreme f(x,y)=3x^3-12x+2y^2-8y+7 is Saddle(-2/(sqrt(3)),2),Minimum(2/(sqrt(3)),2)

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extreme f(x,y)=3x^3-12x+2y^2-8y+7

extreme points

f (x, y) = 3 x^{3} - 12 x + 2 y^{2} - 8 y + 7

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 72 x

Check the sign of at each critical point

Check critical point

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

Saddle

Check critical point

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

Minimum

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

f (x, y) = (\frac{1}{16 - x ^{2} - y ^{2}})

6 x

(210 - x) (110 + x) - 24 (210 - x)

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

f (x) = 12 x^{2} - 48 x + 20

What is the extreme f(x,y)=3x^3-12x+2y^2-8y+7 ?
The extreme f(x,y)=3x^3-12x+2y^2-8y+7 is Saddle(-2/(sqrt(3)),2),Minimum(2/(sqrt(3)),2)