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

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

extreme points

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

Minimum (0, 0)

Solution steps

f^{'} (x) = - \frac{4 x}{( x - 1 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

y = 5 x^{2} + 6 x + 9

6 e^{x + 1} - 3

\frac{ln ( x )}{x}

\frac{x + 5}{x ^{2} - 9}

lo g_{4} (x + 2) - 2 lo g_{4} (1 - x) + 1

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