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

The extreme f(x,y)=(x-y)(2x+y)(x+1) is Saddle(0,0),Maximum(-2/3 , 1/3)

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-y)(2x+y)(x+1)

extreme points

f (x, y) = (x - y) (2 x + y) (x + 1)

Saddle (0, 0), Maximum (- \frac{2}{3}, \frac{1}{3})

Solution steps

Find the critical points:

(0, 0), (- \frac{2}{3}, \frac{1}{3})

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

D (x, y) = - 28 x^{2} - 4 x y - 36 x - 4 y^{2} - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{2}{3}, \frac{1}{3}) :

Maximum

Saddle (0, 0), Maximum (- \frac{2}{3}, \frac{1}{3})

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

2 x^{2} + 9

y = 2 x^{4} - 12 x^{2}

f (x) = \frac{x ^{2} + 2}{x ^{2} + 3}

- x^{2} - x + 1

What is the extreme f(x,y)=(x-y)(2x+y)(x+1) ?
The extreme f(x,y)=(x-y)(2x+y)(x+1) is Saddle(0,0),Maximum(-2/3 , 1/3)