What is the extreme f(x)=(4x)/(3x^2+27) ?

The extreme f(x)=(4x)/(3x^2+27) is Minimum(-3,-2/9),Maximum(3, 2/9)

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extreme f(x)=(4x)/(3x^2+27)

extreme points

f (x) = \frac{4 x}{3 x ^{2} + 27}

Minimum (- 3, - \frac{2}{9}), Maximum (3, \frac{2}{9})

Solution steps

f^{'} (x) = - \frac{4 ( x ^{2} - 9 )}{3 ( 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}{3 x ^{2} + 27} : - \frac{2}{9}

Minimum (- 3, - \frac{2}{9})

Plug

x = 3

into

\frac{4 x}{3 x ^{2} + 27} : \frac{2}{9}

Maximum (3, \frac{2}{9})

Minimum (- 3, - \frac{2}{9}), Maximum (3, \frac{2}{9})

f (x) = \frac{1}{3} x^{3} + 2 x^{2} + 4 x

x^{5} - 5 x^{3}

y = \frac{- 2 x}{9 + x ^{2}}

f (x) = - x^{2} - 12 x - 27

f (x) = x 9 - x^{2}, - 3 \leq x \leq 3

What is the extreme f(x)=(4x)/(3x^2+27) ?
The extreme f(x)=(4x)/(3x^2+27) is Minimum(-3,-2/9),Maximum(3, 2/9)