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

The extreme f(x,y)=(x^3-y^3)/(x-y) is Minimum(0,0)

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-y^3)/(x-y)

extreme points

f (x, y) = \frac{x ^{3} - y ^{3}}{x - y}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = x^{4} - 98 x^{2} - 1, (- 8, 8)

f (x, y) = (x - 1)^{2} + (y - 4)^{2}

5 x + \frac{100}{x}

f (x) = 3 x^{3} - 9 x^{2} - 315 x + 7

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

What is the extreme f(x,y)=(x^3-y^3)/(x-y) ?
The extreme f(x,y)=(x^3-y^3)/(x-y) is Minimum(0,0)