What is the extreme f(x,y)=x^2+y^2-10 ?

The extreme f(x,y)=x^2+y^2-10 is Minimum(0,0)

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extreme f(x,y)=x^2+y^2-10

extreme points

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

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)

f (x) = \frac{( x + 1 )}{x ^{2} + x + 1}

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

\frac{ln ( x )}{x}

f (x) = x, 0 \leq x \leq 16

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

What is the extreme f(x,y)=x^2+y^2-10 ?
The extreme f(x,y)=x^2+y^2-10 is Minimum(0,0)