What is the extreme f(x)=x^2+2y+y^2-8 ?

The extreme f(x)=x^2+2y+y^2-8 is Minimum(0,-1)

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extreme f(x)=x^2+2y+y^2-8

extreme points

f (x) = x^{2} + 2 y + y^{2} - 8

Minimum (0, - 1)

Solution steps

Find the critical points:

(0, - 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, - 1) :

Minimum

Minimum (0, - 1)

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

y = cos (3 x) - 2

f (x) = - 3 x^{3} + 4 x^{2}, - 2 \leq x \leq 2

\frac{x ^{2} + 36}{x ^{2}}

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

What is the extreme f(x)=x^2+2y+y^2-8 ?
The extreme f(x)=x^2+2y+y^2-8 is Minimum(0,-1)