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

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

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

extreme points

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

Saddle (0, - \frac{4}{3})

Solution steps

Find the critical points:

(0, - \frac{4}{3})

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

D (x, y) = - 12 x^{2} y^{2} - 24 x y - 9

Check the sign of at each critical point

Check critical point

(0, - \frac{4}{3}) :

Saddle

Saddle (0, - \frac{4}{3})

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

x^{2} + 2 x + 5

f (x) = 420 t - 35 t

f (x) = x^{2} - 25

\frac{x ^{2}}{x + 2}

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