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^{4} - 8 x^{2} + 1

f (x) = x^{3} - 12 x + 17

f (x) = x^{3} + 2 x^{2} - 20 x

e^{4 x} + e^{- x}

f (x) = x^{2} - 12 x + 8

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)