What is the extreme f(x)=(-4x)/(x^2+6),(0,6) ?

The extreme f(x)=(-4x)/(x^2+6),(0,6) is Minimum(sqrt(6),-sqrt(2/3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=(-4x)/(x^2+6),(0,6)

extreme

Minimum (6, - \frac{2}{3})

Solution steps

Find the critical points:

x = 6

Domain of

\frac{- 4 x}{x ^{2} + 6}, (0, 6) : 0 < x < 6

Find intervals:

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < 6

Plug the extreme point

x = 6

into

\frac{- 4 x}{x ^{2} + 6} \Rightarrow y = - \frac{2}{3}

Minimum (6, - \frac{2}{3})

What is the extreme f(x)=(-4x)/(x^2+6),(0,6) ?
The extreme f(x)=(-4x)/(x^2+6),(0,6) is Minimum(sqrt(6),-sqrt(2/3))