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))

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

extreme points

f (x) = \frac{- 4 x}{x ^{2} + 6}, (0, 6)

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})

f (x) = 2 x^{3} + 15 x^{2} - 36 x + 7, - 6 < x < 2

f (x) = 13 e^{- x}, (0, 4)

- y^{2} + 120 y

f (x) = x - \frac{1}{x} + \frac{2}{x ^{3}}, [0, \infty]

y = 2 x_{1} + 4 x_{2}

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))