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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^3-3x+y^2

extreme points

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

Saddle (- 1, 0), Minimum (1, 0)

Solution steps

Find the critical points:

(- 1, 0), (1, 0)

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

D (x, y) = 12 x

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Saddle

Check critical point

(1, 0) :

Minimum

Saddle (- 1, 0), Minimum (1, 0)

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