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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

Minimum (0, \frac{1}{2})

Solution steps

Find the critical points:

(0, \frac{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

(0, \frac{1}{2}) :

Minimum

Minimum (0, \frac{1}{2})

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