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

The extreme f(x)=4x^2-xy+2y^2 is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 31

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = \frac{- 3 x + y - 4}{4 x ^{2} - 3 y + 2}

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

a - \frac{ab}{( a + b )}

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

f (x) = x^{2} + 9, - 1 \leq x \leq 4

What is the extreme f(x)=4x^2-xy+2y^2 ?
The extreme f(x)=4x^2-xy+2y^2 is Minimum(0,0)