What is the extreme f(x,y)=6x^2+2y^2 ?

The extreme f(x,y)=6x^2+2y^2 is Minimum(0,0)

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extreme f(x,y)=6x^2+2y^2

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 48

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = \frac{2 x - 5}{x}

\frac{5}{x + 6}

f (x) = x^{5} - 5 x^{4} - x^{3} + 28 x^{2} - 2 x

f (x) = (x - 1) (x + 2)

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

What is the extreme f(x,y)=6x^2+2y^2 ?
The extreme f(x,y)=6x^2+2y^2 is Minimum(0,0)