What is the extreme f(x,y)=x^3+y^2+3/2 x^2-6x-4y+3 ?

The extreme f(x,y)=x^3+y^2+3/2 x^2-6x-4y+3 is Minimum(1,2),Saddle(-2,2)

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extreme f(x,y)=x^3+y^2+3/2 x^2-6x-4y+3

extreme points

f (x, y) = x^{3} + y^{2} + \frac{3}{2} x^{2} - 6 x - 4 y + 3

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

Solution steps

Find the critical points:

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

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

D (x, y) = 2 (6 x + 3)

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Check critical point

(- 2, 2) :

Saddle

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

f (x) = 2 x^{3} + 3 x^{2} - 36 x + 5

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

f (x) = - ln (\frac{1}{x + y})

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

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

What is the extreme f(x,y)=x^3+y^2+3/2 x^2-6x-4y+3 ?
The extreme f(x,y)=x^3+y^2+3/2 x^2-6x-4y+3 is Minimum(1,2),Saddle(-2,2)