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

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

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

extreme points

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

Minimum (2, 0), Saddle (2, 2)

Solution steps

Find the critical points:

(2, 0), (2, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6 y + 6

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

Check the sign of at each critical point

Check critical point

(2, 0) :

Minimum

Check critical point

(2, 2) :

Saddle

Minimum (2, 0), Saddle (2, 2)

f (x) = 0.4 x^{2} + 110.2 x, 0 \leq x

f (x) = \frac{( 3 x )}{( 9 - x ^{2} )}, - 3 \leq x \leq 2

f (x) = \frac{1}{4} x + \frac{1}{x}, x > 0

f (0) = x^{2} - 4 x - 4

\frac{7 e ^{x} + 7 e ^{- x}}{2}

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