What is the extreme f(x)= x/((x^2+9)) ?

The extreme f(x)= x/((x^2+9)) is Minimum(-3,-1/6),Maximum(3, 1/6)

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extreme f(x)= x/((x^2+9))

extreme

Minimum (- 3, - \frac{1}{6}), Maximum (3, \frac{1}{6})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 9}{( x ^{2} + 9 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{x}{x ^{2} + 9} : - \frac{1}{6}

Minimum (- 3, - \frac{1}{6})

Plug

x = 3

into

\frac{x}{x ^{2} + 9} : \frac{1}{6}

Maximum (3, \frac{1}{6})

Minimum (- 3, - \frac{1}{6}), Maximum (3, \frac{1}{6})

What is the extreme f(x)= x/((x^2+9)) ?
The extreme f(x)= x/((x^2+9)) is Minimum(-3,-1/6),Maximum(3, 1/6)