What is the extreme f(x,y)=x^2+y^2+5 ?

The extreme f(x,y)=x^2+y^2+5 is Minimum(0,0)

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extreme f(x,y)=x^2+y^2+5

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

f (x, y) = 32 x^{4} + y^{2} - 16 x y

f (x) = x^{3} - 6 x y + y^{3}

f (x) = 2 x^{3} - 3 x^{2} y - 12 x^{2} - 3 y^{2}

f (x) = \frac{x ^{2} + 9}{2 x}

What is the extreme f(x,y)=x^2+y^2+5 ?
The extreme f(x,y)=x^2+y^2+5 is Minimum(0,0)