What is the extreme f(x,y)=(2x^2+4y^2+1)/2 ?

The extreme f(x,y)=(2x^2+4y^2+1)/2 is Minimum(0,0)

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extreme f(x,y)=(2x^2+4y^2+1)/2

extreme points

f (x, y) = \frac{2 x ^{2} + 4 y ^{2} + 1}{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

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

f (x, y) = 4 x + 5 y

6 x^{2} - 2 x^{4}

x - 2 ln (x + 1)

What is the extreme f(x,y)=(2x^2+4y^2+1)/2 ?
The extreme f(x,y)=(2x^2+4y^2+1)/2 is Minimum(0,0)