What is the extreme f(x)=(6x)/(x^2+1) ?

The extreme f(x)=(6x)/(x^2+1) is Minimum(-1,-3),Maximum(1,3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(6x)/(x^2+1)

extreme points

f (x) = \frac{6 x}{x ^{2} + 1}

Minimum (- 1, - 3), Maximum (1, 3)

Solution steps

f^{'} (x) = \frac{6 ( - 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{6 x}{x ^{2} + 1} : - 3

Minimum (- 1, - 3)

Plug

x = 1

into

\frac{6 x}{x ^{2} + 1} : 3

Maximum (1, 3)

Minimum (- 1, - 3), Maximum (1, 3)

(6, 19.1), (7, 19.8)

f (x) = - lo g_{4} (x + 4) - 6

f (x) = e^{\frac{1}{x}}

\frac{x}{x - 2}

\frac{x}{ln ( x )}

What is the extreme f(x)=(6x)/(x^2+1) ?
The extreme f(x)=(6x)/(x^2+1) is Minimum(-1,-3),Maximum(1,3)