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

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

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

extreme points

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

Minimum (0, 0), Saddle (4, - 4)

Solution steps

Find the critical points:

(0, 0), (4, - 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = - 48 x + 96

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(4, - 4) :

Saddle

Minimum (0, 0), Saddle (4, - 4)

200 e^{- 0.1 p}

- x^{3} + 2

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

y^{2} - x y - x^{2}

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

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