What is the extreme f(x,y)=14x^2-2x^3+4xy ?

The extreme f(x,y)=14x^2-2x^3+4xy is Saddle(0,0)

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extreme f(x,y)=14x^2-2x^3+4xy

extreme points

f (x, y) = 14 x^{2} - 2 x^{3} + 4 x y

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = \frac{2}{3} x^{3} + 2 x^{2} + 6 x - 5

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

f (x) = 2 cos^{2} (x)

6 x^{\frac{2}{3}} - 4 x

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

What is the extreme f(x,y)=14x^2-2x^3+4xy ?
The extreme f(x,y)=14x^2-2x^3+4xy is Saddle(0,0)