What is the extreme f(x)= x/(x^2+36) ?

The extreme f(x)= x/(x^2+36) is Minimum(-6,-1/12),Maximum(6, 1/12)

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extreme f(x)= x/(x^2+36)

extreme points

f (x) = \frac{x}{x ^{2} + 36}

Minimum (- 6, - \frac{1}{12}), Maximum (6, \frac{1}{12})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 36}{( x ^{2} + 36 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 6,

Increasing

: - 6 < x < 6,

Decreasing

: 6 < x < \infty

Plug

x = - 6

into

\frac{x}{x ^{2} + 36} : - \frac{1}{12}

Minimum (- 6, - \frac{1}{12})

Plug

x = 6

into

\frac{x}{x ^{2} + 36} : \frac{1}{12}

Maximum (6, \frac{1}{12})

Minimum (- 6, - \frac{1}{12}), Maximum (6, \frac{1}{12})

What is the extreme f(x)= x/(x^2+36) ?
The extreme f(x)= x/(x^2+36) is Minimum(-6,-1/12),Maximum(6, 1/12)