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 points

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

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})

f (x) = \frac{x ^{2}}{x - 3}

1 + x

x 4 - x^{2}

\frac{2 x ^{2} + 16 x - 18}{x ^{2} + x - 6}

f (t) = \frac{3}{x - 4}

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)