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

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

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

extreme points

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

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)

f (x) = (x - 1)^{2}

32 x y

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

f (x) = x^{3} - x^{2} - 5 x + 7

w (x, y) = x^{2} + y^{2}

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