What is the extreme f(x)=(5x)/(x^2+9) ?

The extreme f(x)=(5x)/(x^2+9) is Minimum(-3,-5/6),Maximum(3, 5/6)

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extreme f(x)=(5x)/(x^2+9)

extreme points

f (x) = \frac{5 x}{x ^{2} + 9}

Minimum (- 3, - \frac{5}{6}), Maximum (3, \frac{5}{6})

Solution steps

f^{'} (x) = \frac{5 ( - 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{5 x}{x ^{2} + 9} : - \frac{5}{6}

Minimum (- 3, - \frac{5}{6})

Plug

x = 3

into

\frac{5 x}{x ^{2} + 9} : \frac{5}{6}

Maximum (3, \frac{5}{6})

Minimum (- 3, - \frac{5}{6}), Maximum (3, \frac{5}{6})

x^{9} - 9 x

x^{2} + 4 x - 5

f (x) = 2 sin (x) - x, 0 \leq x \leq 2 π

f (x) = 8 + 6 x - 7 x^{2} - 4 x^{3}

f (x) = \frac{20}{3 ( 2 + x ) ^{3}}

What is the extreme f(x)=(5x)/(x^2+9) ?
The extreme f(x)=(5x)/(x^2+9) is Minimum(-3,-5/6),Maximum(3, 5/6)