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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(- 1, 0) :

Saddle

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

(x^{2} - 3 x + 5) e^{- \frac{x}{3}}

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

f (x) = y^{3} + 3 x^{2} y - 6 x^{2} - 6 y^{2} + 2

f (x) = \frac{e ^{x}}{1 - e ^{x}}

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

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