What is the extreme f(x,y)=xy^2-6x^2-3x+2xy+9 ?

The extreme f(x,y)=xy^2-6x^2-3x+2xy+9 is Saddle(0,1),Saddle(0,-3),Maximum(-1/3 ,-1)

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extreme f(x,y)=xy^2-6x^2-3x+2xy+9

extreme points

f (x, y) = x y^{2} - 6 x^{2} - 3 x + 2 x y + 9

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 24 x - 4 y^{2} - 8 y - 4

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 3) :

Saddle

Check critical point

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

Maximum

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

f (p, q) = pq - \frac{1}{p} - \frac{1}{q}

3 x^{4} - 4 x^{3} - 36 x^{2} + 5

f (x) = x 18 - x

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

f (x) = 5 sin (2 x) + 3

What is the extreme f(x,y)=xy^2-6x^2-3x+2xy+9 ?
The extreme f(x,y)=xy^2-6x^2-3x+2xy+9 is Saddle(0,1),Saddle(0,-3),Maximum(-1/3 ,-1)