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

The extreme f(x)=x^2+xy+y^2-y is Minimum(-1/3 , 2/3)

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

extreme points

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

Minimum (- \frac{1}{3}, \frac{2}{3})

Solution steps

Find the critical points:

(- \frac{1}{3}, \frac{2}{3})

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- \frac{1}{3}, \frac{2}{3}) :

Minimum

Minimum (- \frac{1}{3}, \frac{2}{3})

f (x) = x^{2} - 4 x + 1, 0 \leq x \leq 3

x - 3

f (x, y) = 3 y^{2} + x^{3} - 6 x y - 9 x

f (x) = 5 sin (x) cos (x)

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

What is the extreme f(x)=x^2+xy+y^2-y ?
The extreme f(x)=x^2+xy+y^2-y is Minimum(-1/3 , 2/3)