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

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

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

extreme points

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

Minimum (- 1, 0)

Solution steps

Find the critical points:

(- 1, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Minimum

Minimum (- 1, 0)

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

V (B, h) = \frac{1}{3} B h

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

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

f (x) = \frac{1}{x} + ln (x)

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