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)

\frac{4 x}{x ^{2} - 25}

f (x) = x^{2} + x y + y^{2} - 10 y + 33

f (x) = x^{3} - 6 x^{2} + 12 x + 1

f (x) = 1 - 2 x^{2} - y^{2}

f (x) = e^{5 x} + e^{- x}

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)