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

The extreme f(x)=x^2sqrt(4-x) is Minimum(0,0),Maximum(16/5 ,(512)/(25sqrt(5))),Minimum(4,0)

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

extreme points

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

Minimum (0, 0), Maximum (\frac{16}{5}, \frac{512}{25 5}), Minimum (4, 0)

Solution steps

f^{'} (x) = \frac{x ( - 5 x + 16 )}{2 4 - x}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{16}{5},

Decreasing

: \frac{16}{5} < x < 4

Plug

x = 0

into

x^{2} 4 - x : 0

Minimum (0, 0)

Plug

x = \frac{16}{5}

into

x^{2} 4 - x : \frac{512}{25 5}

Maximum (\frac{16}{5}, \frac{512}{25 5})

Plug

x = 4

into

x^{2} 4 - x : 0

Minimum (4, 0)

Minimum (0, 0), Maximum (\frac{16}{5}, \frac{512}{25 5}), Minimum (4, 0)

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

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

f (x, y) = 2 x^{3} + 4 y^{3} + 3 x^{2} - 12 x - 192 y + 5

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

f (x, y) = x^{4} - y^{4} + 2 x^{2} y - 2 y^{2} - 5

What is the extreme f(x)=x^2sqrt(4-x) ?
The extreme f(x)=x^2sqrt(4-x) is Minimum(0,0),Maximum(16/5 ,(512)/(25sqrt(5))),Minimum(4,0)