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

The extreme f(x,y)=-x^2-y^2-1/2 x is Maximum(-1/4 ,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

f (x, y) = - x^{2} - y^{2} - \frac{1}{2} x

Maximum (- \frac{1}{4}, 0)

Solution steps

Find the critical points:

(- \frac{1}{4}, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- \frac{1}{4}, 0) :

Maximum

Maximum (- \frac{1}{4}, 0)

f (x) = \frac{- 2}{x ^{2} - 1}

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

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

x^{3} - 6 x^{2}

f (x) = (x^{2} - 5) (x - 1)^{2} (x - 2)^{3}

What is the extreme f(x,y)=-x^2-y^2-1/2 x ?
The extreme f(x,y)=-x^2-y^2-1/2 x is Maximum(-1/4 ,0)