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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme points

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

Minimum (- 1, - \frac{1}{2}), Maximum (1, \frac{1}{2})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, - \frac{1}{2})

Plug

x = 1

into

\frac{x}{x ^{2} + 1} : \frac{1}{2}

Maximum (1, \frac{1}{2})

Minimum (- 1, - \frac{1}{2}), Maximum (1, \frac{1}{2})

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