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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = \frac{2 ( - 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{2 x}{x ^{2} + 9} : - \frac{1}{3}

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

Plug

x = 3

into

\frac{2 x}{x ^{2} + 9} : \frac{1}{3}

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

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

2 x^{3} - 4 x^{2} + 2 x + 1

y (x, t) = x (3 t)

f (x) = 3 x^{\frac{2}{3}} - 2 x [- 1.1]

f (x) = x^{2} + 3 x + 5

f (x, y) = \frac{1}{3} 36 - 4 x^{2} - 9 y^{2}

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