What is the extreme f(x,y)=x-1/2 y^2-1/3 x^3+y+6 ?

The extreme f(x,y)=x-1/2 y^2-1/3 x^3+y+6 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-1/2 y^2-1/3 x^3+y+6

extreme points

f (x, y) = x - \frac{1}{2} y^{2} - \frac{1}{3} x^{3} + y + 6

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

Solution steps

Find the critical points:

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

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

D (x, y) = 2 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) = \frac{4 x ^{2}}{2 x - 8}

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

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

f (x) = 8 x^{2} - 4 x^{4}

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

What is the extreme f(x,y)=x-1/2 y^2-1/3 x^3+y+6 ?
The extreme f(x,y)=x-1/2 y^2-1/3 x^3+y+6 is Saddle(-1,1),Maximum(1,1)