What is the extreme f(x)=(4x)/(x^2+1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(4x)/(x^2+1)

extreme points

f (x) = \frac{4 x}{x ^{2} + 1}

Minimum (- 1, - 2), Maximum (1, 2)

Solution steps

f^{'} (x) = \frac{4 ( - 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{4 x}{x ^{2} + 1} : - 2

Minimum (- 1, - 2)

Plug

x = 1

into

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

Maximum (1, 2)

Minimum (- 1, - 2), Maximum (1, 2)

f (x) = \frac{x ^{3} - 27}{x ^{2} - 8 x + 15}

e^{x + 2}

x^{2} + 4

y = f (x) = ln (2 x + 1)

(2, - 2), (- 4, 4)

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