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

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

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

extreme points

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

Minimum (\frac{1}{2}, 0)

Solution steps

Find the critical points:

(\frac{1}{2}, 0)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, 0) :

Minimum

Minimum (\frac{1}{2}, 0)

x^{4} - x^{3}

x^{6} (x - 4)^{5}

2 x^{2} + 3

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

f (x) = x^{2} - 8 x + 20

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