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

The extreme f(x,y)=4xy^2+2xy-3y is Saddle(3/2 ,0),Saddle(-3/2 ,-1/2)

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

extreme points

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

Saddle (\frac{3}{2}, 0), Saddle (- \frac{3}{2}, - \frac{1}{2})

Solution steps

Find the critical points:

(\frac{3}{2}, 0), (- \frac{3}{2}, - \frac{1}{2})

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

D (x, y) = - 64 y^{2} - 32 y - 4

Check the sign of at each critical point

Check critical point

(\frac{3}{2}, 0) :

Saddle

Check critical point

(- \frac{3}{2}, - \frac{1}{2}) :

Saddle

Saddle (\frac{3}{2}, 0), Saddle (- \frac{3}{2}, - \frac{1}{2})

f (x) = - \frac{1}{6} x^{3} + \frac{3}{2} x^{2} + \frac{5}{2} x - \frac{11}{6}

f (x) = 3 x - t

f (x, y) = 3 x - 4 y + 5 (1, 1)

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

What is the extreme f(x,y)=4xy^2+2xy-3y ?
The extreme f(x,y)=4xy^2+2xy-3y is Saddle(3/2 ,0),Saddle(-3/2 ,-1/2)