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

The extreme f(x)=4xy^2-2x^2y-x is Saddle(0, 1/2),Saddle(0,-1/2)

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

extreme points

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

Saddle (0, \frac{1}{2}), Saddle (0, - \frac{1}{2})

Solution steps

Find the critical points:

(0, \frac{1}{2}), (0, - \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 8 x

D (x, y) = - 16 x^{2} + 32 x y - 64 y^{2}

Check the sign of at each critical point

Check critical point

(0, \frac{1}{2}) :

Saddle

Check critical point

(0, - \frac{1}{2}) :

Saddle

Saddle (0, \frac{1}{2}), Saddle (0, - \frac{1}{2})

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

f (x) = x^{2} - 2 x + 5

\frac{( x + 4 ) ( x - 1 )}{3 x + 2}

f (x) = x - 5 ln (x)

f (x) = 4 cos (3 x) + 5

What is the extreme f(x)=4xy^2-2x^2y-x ?
The extreme f(x)=4xy^2-2x^2y-x is Saddle(0, 1/2),Saddle(0,-1/2)