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

The extreme f(x,y)=2x^3+6xy+3y^2 is Minimum(1,-1),Saddle(0,0)

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extreme f(x,y)=2x^3+6xy+3y^2

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 72 x - 36

Check the sign of at each critical point

Check critical point

(1, - 1) :

Minimum

Check critical point

(0, 0) :

Saddle

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

f (x, y) = 16 x 3 y - 2 x^{2} - 4 x - 16 y + 2

f (x) = cos (5 x), - \frac{π}{6} \leq x \leq \frac{π}{4}

f (x) = 12 x^{4} - 4 x^{2} + 6

f (x) = 16 sin (2 x), - 2 π \leq x \leq 2 π

f (x) = \frac{x ^{2} + 10 x + 16}{x ^{2} + 4 x + 4}

What is the extreme f(x,y)=2x^3+6xy+3y^2 ?
The extreme f(x,y)=2x^3+6xy+3y^2 is Minimum(1,-1),Saddle(0,0)