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

The extreme f(x)= x/(x^2-x+9) is Minimum(-3,-1/7),Maximum(3, 1/5)

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

extreme points

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

Minimum (- 3, - \frac{1}{7}), Maximum (3, \frac{1}{5})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{x}{x ^{2} - x + 9} : - \frac{1}{7}

Minimum (- 3, - \frac{1}{7})

Plug

x = 3

into

\frac{x}{x ^{2} - x + 9} : \frac{1}{5}

Maximum (3, \frac{1}{5})

Minimum (- 3, - \frac{1}{7}), Maximum (3, \frac{1}{5})

What is the extreme f(x)= x/(x^2-x+9) ?
The extreme f(x)= x/(x^2-x+9) is Minimum(-3,-1/7),Maximum(3, 1/5)