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

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

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

extreme points

f (z) = 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) = 2 x^{3} - 3 x^{2} - 36 x + 54

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

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 1, - 1 \leq x \leq 2

y = 5 x^{2} ln (\frac{x}{4})

\frac{x ^{2} - 8}{x + 3}