What is the extreme f(x,y)=3x^2-6x+4y^2+8 ?

The extreme f(x,y)=3x^2-6x+4y^2+8 is Minimum(1,0)

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extreme f(x,y)=3x^2-6x+4y^2+8

extreme points

f (x, y) = 3 x^{2} - 6 x + 4 y^{2} + 8

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 48

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

f (x) = 2 x^{3} + 11 x^{2} - 8 x - 12

f (x) = 4 x^{2} + 1 - 3 x

f (x) = 75 - x^{2}

f (x) = \frac{x ^{3}}{3} - 3 x^{2} + 8 x - 2, 0 \leq x \leq 3

f (x, y) = 2 x^{3} - 6 x y + 3 y^{2}

What is the extreme f(x,y)=3x^2-6x+4y^2+8 ?
The extreme f(x,y)=3x^2-6x+4y^2+8 is Minimum(1,0)