What is the extreme f(x,y)=4x^2+2y^2+5 ?

The extreme f(x,y)=4x^2+2y^2+5 is Minimum(0,0)

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extreme f(x,y)=4x^2+2y^2+5

extreme points

f (x, y) = 4 x^{2} + 2 y^{2} + 5

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 32

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

1 - ∣ x ∣

f (x) = - 8 x^{3} + 24 x + 9

f (x) = 5 x^{2} - 4 y^{2} - 7 x + 5 y + 4

(3 x + 2) (2 x + 5)

f (x) = - \frac{4}{x - 5}

What is the extreme f(x,y)=4x^2+2y^2+5 ?
The extreme f(x,y)=4x^2+2y^2+5 is Minimum(0,0)