What is the extreme f(x,y)=(x+1)^2+x^2y^2 ?

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

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extreme f(x,y)=(x+1)^2+x^2y^2

extreme points

f (x, y) = (x + 1)^{2} + x^{2} y^{2}

Minimum (- 1, 0)

Solution steps

Find the critical points:

(- 1, 0)

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

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

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Minimum

Minimum (- 1, 0)

f (x, y) = 5 x^{2} + 5 y^{2} + 3

f (x) = \frac{x - 2}{x + 2}

x + \frac{25}{x}

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

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

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