What is the extreme f(x)=4x^2+2y^2-2xy-10y-2x ?

The extreme f(x)=4x^2+2y^2-2xy-10y-2x is Minimum(1,3)

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extreme f(x)=4x^2+2y^2-2xy-10y-2x

extreme points

f (x) = 4 x^{2} + 2 y^{2} - 2 x y - 10 y - 2 x

Minimum (1, 3)

Solution steps

Find the critical points:

(1, 3)

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

D (x, y) = 28

Check the sign of at each critical point

Check critical point

(1, 3) :

Minimum

Minimum (1, 3)

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

f (x) = 2 x^{2} - 6 x^{4}

f (x) = - x^{2} + 10, - 3 \leq x \leq 4

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

\frac{2}{x + 5}

What is the extreme f(x)=4x^2+2y^2-2xy-10y-2x ?
The extreme f(x)=4x^2+2y^2-2xy-10y-2x is Minimum(1,3)