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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Minimum (- 1, - 1), Maximum (1, 1)

Solution steps

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

Minimum (- 1, - 1)

Plug

x = 1

into

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

Maximum (1, 1)

Minimum (- 1, - 1), Maximum (1, 1)

\frac{6}{( t - 8 )}

\frac{- x + 8}{3}

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

f (x) = (8 - 10 x)^{\frac{7}{2}}

f (x) = \frac{x ^{4}}{x ^{2} + 6}

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