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

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

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

extreme points

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

Minimum (1, 2)

Solution steps

Find the critical points:

(1, 2)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Minimum (1, 2)

3 x^{2} - 24 ln (x)

f (x) = \frac{4 x}{x ^{2} + 1}, - 4 \leq x \leq 0

f (x) = ln (27 + x^{3})

f (x) = x^{3} - 9 x^{2} + 27 x - 3

f (x, y) = y^{2} + x y + 3 x + 2 y + 5

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