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

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

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

extreme points

f (x) = 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)

\frac{24}{r} + 2 π r^{2}

\frac{d}{d x} (9 x^{3} y^{5})

f (x) = 6 y^{2} + 4 x^{2} - 12 x y - 4 y

f (x) = x^{3} + 4 x^{2} - 3 x + 6

f (x, y) = x^{3} + y^{3} + 2 x^{2} + 4 y^{2} + 6