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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = 2

into

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

Minimum (2, 4)

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

f (x) = x [18 - 2 (x)]^{2}

f (x, y) = x^{2} - 3 y^{2} - 8 x + 9 y + 3 x y

f (d, y) = \frac{d}{d ^{2}} (y)

y = - 5 cos (3 x) - 2

f (x, y) = 3 x^{\frac{1}{3}} y^{\frac{1}{3}}

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