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

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

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extreme f(x,y)=1+x^2+y^2

extreme points

f (x, y) = 1 + x^{2} + y^{2}

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)

p (a, c) = a + 8 + c

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

f (x) = \frac{x - 1}{x ^{2} + 3 x}

e^{3 x^{2} - x - 3}

f (x) = - 2 x^{3} + 18 x^{2} - 30 x, 0 \leq x \leq 6

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