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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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)

x^{4} - x^{3} + 2 x^{2} - 1

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

f (x, y) = x^{3} - 12 x y + y^{3}

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

f (x) = 3 x ln (x)

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)