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)

f (x) = \frac{30 x}{x - 2}

f (x) = ∣ x + 3 ∣

f (x, y) = 4 x - 3 y

f (x) = 8 + 81 x - 3 x^{3}, 0 \leq x \leq 4

f (x) = x^{4} - 14 x^{2} + 49, 0 \leq x \leq 5

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)