What is the extreme f(x)= x/(x^2+108) ?

The extreme f(x)= x/(x^2+108) is Minimum(-6sqrt(3),-1/(12sqrt(3))),Maximum(6sqrt(3), 1/(12sqrt(3)))

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extreme f(x)= x/(x^2+108)

extreme points

f (x) = \frac{x}{x ^{2} + 108}

Minimum (- 63, - \frac{1}{12 3}), Maximum (63, \frac{1}{12 3})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 108}{( x ^{2} + 108 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 63,

Increasing

: - 63 < x < 63,

Decreasing

: 63 < x < \infty

Plug

x = - 63

into

\frac{x}{x ^{2} + 108} : - \frac{1}{12 3}

Minimum (- 63, - \frac{1}{12 3})

Plug

x = 63

into

\frac{x}{x ^{2} + 108} : \frac{1}{12 3}

Maximum (63, \frac{1}{12 3})

Minimum (- 63, - \frac{1}{12 3}), Maximum (63, \frac{1}{12 3})

f (x) = - 4 x^{2} + 96 x - 490

f (x) = 4 + x + x^{2}

x y^{2} + \frac{8}{x y} - ln (y)

f (x) = 6 x - 6, - 6 \leq x \leq 3

What is the extreme f(x)= x/(x^2+108) ?
The extreme f(x)= x/(x^2+108) is Minimum(-6sqrt(3),-1/(12sqrt(3))),Maximum(6sqrt(3), 1/(12sqrt(3)))