What is the extreme f(x)= x/(x^2+9x+20) ?

The extreme f(x)= x/(x^2+9x+20) is Minimum(-2sqrt(5),9+4sqrt(5)),Maximum(2sqrt(5),9-4sqrt(5))

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+9x+20)

extreme points

f (x) = \frac{x}{x ^{2} + 9 x + 20}

Minimum (- 25, 9 + 45), Maximum (25, 9 - 45)

Solution steps

f^{'} (x) = \frac{- x ^{2} + 20}{( x ^{2} + 9 x + 20 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < - 25,

Increasing

: - 25 < x < - 4,

Increasing

: - 4 < x < 25,

Decreasing

: 25 < x < \infty

Plug

x = - 25

into

\frac{x}{x ^{2} + 9 x + 20} : 9 + 45

Minimum (- 25, 9 + 45)

Plug

x = 25

into

\frac{x}{x ^{2} + 9 x + 20} : 9 - 45

Maximum (25, 9 - 45)

Minimum (- 25, 9 + 45), Maximum (25, 9 - 45)

What is the extreme f(x)= x/(x^2+9x+20) ?
The extreme f(x)= x/(x^2+9x+20) is Minimum(-2sqrt(5),9+4sqrt(5)),Maximum(2sqrt(5),9-4sqrt(5))