What is the extreme f(x,y)=10-x^2y+3xy+xy^2 ?

The extreme f(x,y)=10-x^2y+3xy+xy^2 is Minimum(1,-1),Saddle(0,-3)

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extreme f(x,y)=10-x^2y+3xy+xy^2

extreme points

f (x, y) = 10 - x^{2} y + 3 x y + x y^{2}

Minimum (1, - 1), Saddle (0, - 3)

Solution steps

Find the critical points:

(1, - 1), (0, - 3)

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

D (x, y) = 4 x y + 12 x - 4 x^{2} - 4 y^{2} - 12 y - 9

Check the sign of at each critical point

Check critical point

(1, - 1) :

Minimum

Check critical point

(0, - 3) :

Saddle

Minimum (1, - 1), Saddle (0, - 3)

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

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

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

y = x^{3} - 6 x^{2} + 9 x

f (x) = 2 x^{4} - 16 x^{2} + 20 x - 7

What is the extreme f(x,y)=10-x^2y+3xy+xy^2 ?
The extreme f(x,y)=10-x^2y+3xy+xy^2 is Minimum(1,-1),Saddle(0,-3)