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

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

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

extreme points

f (x) = 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) = ln (x^{2} + 7 x + 15)

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

x^{3} - 3 x^{2} + 5

f (x) = \frac{x ^{2}}{4 - x ^{2}}

f (x, y) = \frac{1}{5} x^{5} - \frac{5}{3} x^{3} + 4 x + x y^{2}

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