What is the extreme f(x)=x^2+xy+1/2 y^2-4x+y ?

The extreme f(x)=x^2+xy+1/2 y^2-4x+y is Minimum(5,-6)

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extreme f(x)=x^2+xy+1/2 y^2-4x+y

extreme points

f (x) = x^{2} + x y + \frac{1}{2} y^{2} - 4 x + y

Minimum (5, - 6)

Solution steps

Find the critical points:

(5, - 6)

\frac{\partial ^{2} f}{\partial x ^{2}} = 1

D (x, y) = 1

Check the sign of at each critical point

Check critical point

(5, - 6) :

Minimum

Minimum (5, - 6)

f (x) = \frac{1}{3} x^{3} - 2 x^{2} - 5 x + 2

x y e^{x + 2 y}

1 - 2 x^{2}

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

x^{4} - 8 x^{2} + 5

What is the extreme f(x)=x^2+xy+1/2 y^2-4x+y ?
The extreme f(x)=x^2+xy+1/2 y^2-4x+y is Minimum(5,-6)