What is the extreme f(x,y)=x^3+3xy^2-3y^2+3x^2+11 ?

The extreme f(x,y)=x^3+3xy^2-3y^2+3x^2+11 is Saddle(0,0),Maximum(-2,0)

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extreme f(x,y)=x^3+3xy^2-3y^2+3x^2+11

extreme points

f (x, y) = x^{3} + 3 x y^{2} - 3 y^{2} + 3 x^{2} + 11

Saddle (0, 0), Maximum (- 2, 0)

Solution steps

Find the critical points:

(0, 0), (- 2, 0)

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

D (x, y) = (6 x + 6) (6 x - 6) - 36 y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, 0) :

Maximum

Saddle (0, 0), Maximum (- 2, 0)

f (x) = - \frac{10}{x ^{2} - 3 x + 14}

y = x^{2} + 3

f (x, y) = \frac{1}{3 ^{- x^{2} - y^{2}} + 1}

f (x) = - sin^{2} (x), - π \leq x \leq π

\frac{- 6}{x - 7}

What is the extreme f(x,y)=x^3+3xy^2-3y^2+3x^2+11 ?
The extreme f(x,y)=x^3+3xy^2-3y^2+3x^2+11 is Saddle(0,0),Maximum(-2,0)