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

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

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

extreme points

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

Saddle (0, 0), Minimum (\frac{2}{3}, - \frac{4}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{2}{3}, - \frac{4}{3})

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

D (x, y) = 12 x - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{2}{3}, - \frac{4}{3}) :

Minimum

Saddle (0, 0), Minimum (\frac{2}{3}, - \frac{4}{3})

- sin (2 x)

f (x) = - 2 x^{3} + 42 x^{2} - 270 x + 3

f (x) = - 2 x^{3} + 42 x^{2} - 270 x + 8

x^{4} - y^{4} - 2 x^{2} + 2 y^{2}

f (x) = \frac{x}{8 x - 3}, 8 \leq x \leq 13

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