What is the extreme f(x)=(sqrt(x)(x-4)^2)/4 ?

The extreme f(x)=(sqrt(x)(x-4)^2)/4 is Minimum(0,0),Maximum(4/5 ,(128)/(25sqrt(5))),Minimum(4,0)

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extreme f(x)=(sqrt(x)(x-4)^2)/4

extreme points

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

Minimum (0, 0), Maximum (\frac{4}{5}, \frac{128}{25 5}), Minimum (4, 0)

Solution steps

f^{'} (x) = \frac{5 x ^{2} - 24 x + 16}{8 x}

Find intervals:

Increasing

: 0 < x < \frac{4}{5},

Decreasing

: \frac{4}{5} < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

\frac{x ( x - 4 ) ^{2}}{4} : 0

Minimum (0, 0)

Plug

x = \frac{4}{5}

into

\frac{x ( x - 4 ) ^{2}}{4} : \frac{128}{25 5}

Maximum (\frac{4}{5}, \frac{128}{25 5})

Plug

x = 4

into

\frac{x ( x - 4 ) ^{2}}{4} : 0

Minimum (4, 0)

Minimum (0, 0), Maximum (\frac{4}{5}, \frac{128}{25 5}), Minimum (4, 0)

y (t, β) = - 4 e^{- 4 t} (β + 15) + 5 e^{- 5 t} (β + 12)

f (x, y) = 2 x^{2} + 2 y^{2} - 4 y + 9

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

f (x) = x (x^{3} - x + 10)

f (x) = - 462 x + 64 x^{2} - 2 x^{3}

What is the extreme f(x)=(sqrt(x)(x-4)^2)/4 ?
The extreme f(x)=(sqrt(x)(x-4)^2)/4 is Minimum(0,0),Maximum(4/5 ,(128)/(25sqrt(5))),Minimum(4,0)