What is the extreme f(x)=x^2-2x+6y^2 ?

The extreme f(x)=x^2-2x+6y^2 is Minimum(1,0)

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extreme f(x)=x^2-2x+6y^2

extreme points

f (x) = x^{2} - 2 x + 6 y^{2}

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

f (x) = 4 x + 3

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

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x - 1, 2 \leq x \leq 7

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

f (x) = 416 - x x

What is the extreme f(x)=x^2-2x+6y^2 ?
The extreme f(x)=x^2-2x+6y^2 is Minimum(1,0)