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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

Minimum

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

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

y = x^{3} - 12 x

f (x) = t^{3} - 13 t^{2} + 50 t - 56

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

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

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