What is the extreme f(x)= x/(x^2+81) ?

The extreme f(x)= x/(x^2+81) is Minimum(-9,-1/18),Maximum(9, 1/18)

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extreme f(x)= x/(x^2+81)

extreme points

f (x) = \frac{x}{x ^{2} + 81}

Minimum (- 9, - \frac{1}{18}), Maximum (9, \frac{1}{18})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 81}{( x ^{2} + 81 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 9,

Increasing

: - 9 < x < 9,

Decreasing

: 9 < x < \infty

Plug

x = - 9

into

\frac{x}{x ^{2} + 81} : - \frac{1}{18}

Minimum (- 9, - \frac{1}{18})

Plug

x = 9

into

\frac{x}{x ^{2} + 81} : \frac{1}{18}

Maximum (9, \frac{1}{18})

Minimum (- 9, - \frac{1}{18}), Maximum (9, \frac{1}{18})

What is the extreme f(x)= x/(x^2+81) ?
The extreme f(x)= x/(x^2+81) is Minimum(-9,-1/18),Maximum(9, 1/18)