What is the extreme f(x)=(4x)/(x^2+9) ?

The extreme f(x)=(4x)/(x^2+9) is Minimum(-3,-2/3),Maximum(3, 2/3)

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extreme f(x)=(4x)/(x^2+9)

extreme points

f (x) = \frac{4 x}{x ^{2} + 9}

Minimum (- 3, - \frac{2}{3}), Maximum (3, \frac{2}{3})

Solution steps

f^{'} (x) = \frac{4 ( - 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{4 x}{x ^{2} + 9} : - \frac{2}{3}

Minimum (- 3, - \frac{2}{3})

Plug

x = 3

into

\frac{4 x}{x ^{2} + 9} : \frac{2}{3}

Maximum (3, \frac{2}{3})

Minimum (- 3, - \frac{2}{3}), Maximum (3, \frac{2}{3})

f (x y) = x^{3} + y^{3} - x y

f (x) = x^{2} ln (x), \frac{1}{3} \leq x \leq \frac{5}{6}

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

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

f (x) = t 9 - t^{2}

What is the extreme f(x)=(4x)/(x^2+9) ?
The extreme f(x)=(4x)/(x^2+9) is Minimum(-3,-2/3),Maximum(3, 2/3)