What is the extreme f(x)=2+4x+2x^2-xy+y^2 ?

The extreme f(x)=2+4x+2x^2-xy+y^2 is Minimum(-8/7 ,-4/7)

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extreme f(x)=2+4x+2x^2-xy+y^2

extreme points

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

Minimum (- \frac{8}{7}, - \frac{4}{7})

Solution steps

Find the critical points:

(- \frac{8}{7}, - \frac{4}{7})

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

D (x, y) = 7

Check the sign of at each critical point

Check critical point

(- \frac{8}{7}, - \frac{4}{7}) :

Minimum

Minimum (- \frac{8}{7}, - \frac{4}{7})

f (x) = x e^{- 7 x^{2}}

f (x) = \frac{1}{3} x^{3} - \frac{1}{3} x^{2} - 2 x

f (x) = 3 x^{\frac{5}{3}} - 15 x^{\frac{2}{3}}

f (x) = e^{a x}

f (x, y) = x^{3} + y^{3} - 3 x - 12 y + 20

What is the extreme f(x)=2+4x+2x^2-xy+y^2 ?
The extreme f(x)=2+4x+2x^2-xy+y^2 is Minimum(-8/7 ,-4/7)