What is the extreme f(x)= 1/8 (x+4)^2(6-x) ?

The extreme f(x)= 1/8 (x+4)^2(6-x) is Minimum(-4,0),Maximum(8/3 , 500/27)

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extreme f(x)= 1/8 (x+4)^2(6-x)

extreme points

f (x) = \frac{1}{8} (x + 4)^{2} (6 - x)

Minimum (- 4, 0), Maximum (\frac{8}{3}, \frac{500}{27})

Solution steps

f^{'} (x) = \frac{- 3 x ^{2} - 4 x + 32}{8}

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < \frac{8}{3},

Decreasing

: \frac{8}{3} < x < \infty

Plug

x = - 4

into

\frac{1}{8} (x + 4)^{2} (6 - x) : 0

Minimum (- 4, 0)

Plug

x = \frac{8}{3}

into

\frac{1}{8} (x + 4)^{2} (6 - x) : \frac{500}{27}

Maximum (\frac{8}{3}, \frac{500}{27})

Minimum (- 4, 0), Maximum (\frac{8}{3}, \frac{500}{27})

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

f (x) = - x^{3} + 3 x^{2} + 9 x - 27

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

\frac{( x + 1 )}{x ^{2} - 2 x - 3}

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

What is the extreme f(x)= 1/8 (x+4)^2(6-x) ?
The extreme f(x)= 1/8 (x+4)^2(6-x) is Minimum(-4,0),Maximum(8/3 , 500/27)