What is the extreme f(x)=(x^2)/(1-x) ?

The extreme f(x)=(x^2)/(1-x) is Minimum(0,0),Maximum(2,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=(x^2)/(1-x)

extreme points

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

Minimum (0, 0), Maximum (2, - 4)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{1 - x} : 0

Minimum (0, 0)

Plug

x = 2

into

\frac{x ^{2}}{1 - x} : - 4

Maximum (2, - 4)

Minimum (0, 0), Maximum (2, - 4)

What is the extreme f(x)=(x^2)/(1-x) ?
The extreme f(x)=(x^2)/(1-x) is Minimum(0,0),Maximum(2,-4)