What is the extreme f(x,y)=x^3-4x^2-4y^2+8xy-3x ?

The extreme f(x,y)=x^3-4x^2-4y^2+8xy-3x is Saddle(1,1),Maximum(-1,-1)

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^3-4x^2-4y^2+8xy-3x

extreme points

f (x, y) = x^{3} - 4 x^{2} - 4 y^{2} + 8 x y - 3 x

Saddle (1, 1), Maximum (- 1, - 1)

Solution steps

Find the critical points:

(1, 1), (- 1, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8

D (x, y) = - 48 x

Check the sign of at each critical point

Check critical point

(1, 1) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Saddle (1, 1), Maximum (- 1, - 1)

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

f (x) = x^{3} - 8 x^{2} - 12 x + 1

f (x) = 3 cos (x), 0 \leq x \leq 4 π

4 x + 6 y - x^{2} - y^{2}

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

What is the extreme f(x,y)=x^3-4x^2-4y^2+8xy-3x ?
The extreme f(x,y)=x^3-4x^2-4y^2+8xy-3x is Saddle(1,1),Maximum(-1,-1)