What is the extreme f(x,y)=x+4/(x^2y)+y ?

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

extreme points

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

Minimum (2, 1), Maximum (- 2, - 1)

Solution steps

Find the critical points:

(2, 1), (- 2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{8}{x ^{2} y ^{3}}

D (x, y) = \frac{128}{x ^{6} y ^{4}}

Check the sign of at each critical point

Check critical point

(2, 1) :

Minimum

Check critical point

(- 2, - 1) :

Maximum

Minimum (2, 1), Maximum (- 2, - 1)

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

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

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

y = 2^{x} \cdot I n (5)

f (x) = 3 x^{4} - 4 x^{3} + 5

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