What is the extreme f(x)=x^2+xy+y^2-12x+9 ?

The extreme f(x)=x^2+xy+y^2-12x+9 is Minimum(8,-4)

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extreme f(x)=x^2+xy+y^2-12x+9

extreme points

f (x) = x^{2} + x y + y^{2} - 12 x + 9

Minimum (8, - 4)

Solution steps

Find the critical points:

(8, - 4)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(8, - 4) :

Minimum

Minimum (8, - 4)

f (x) = 5 + (8 + 5 x)^{\frac{2}{5}}

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

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

e^{x} + \frac{1 - x}{e ^{x}}

y = f (x) = x (x - 1)^{5}

What is the extreme f(x)=x^2+xy+y^2-12x+9 ?
The extreme f(x)=x^2+xy+y^2-12x+9 is Minimum(8,-4)