What is the extreme f(x)=x^2sqrt(3-x) ?

The extreme f(x)=x^2sqrt(3-x) is Minimum(0,0),Maximum(12/5 ,(144sqrt(3))/(25sqrt(5))),Minimum(3,0)

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extreme f(x)=x^2sqrt(3-x)

extreme points

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

Minimum (0, 0), Maximum (\frac{12}{5}, \frac{144 3}{25 5}), Minimum (3, 0)

Solution steps

f^{'} (x) = \frac{x ( - 5 x + 12 )}{2 3 - x}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{12}{5},

Decreasing

: \frac{12}{5} < x < 3

Plug

x = 0

into

x^{2} 3 - x : 0

Minimum (0, 0)

Plug

x = \frac{12}{5}

into

x^{2} 3 - x : \frac{144 3}{25 5}

Maximum (\frac{12}{5}, \frac{144 3}{25 5})

Plug

x = 3

into

x^{2} 3 - x : 0

Minimum (3, 0)

Minimum (0, 0), Maximum (\frac{12}{5}, \frac{144 3}{25 5}), Minimum (3, 0)

f (x) = 1 - x

f (x) = - 3 x^{2} - 5 x + 7

2 x^{3} - 2 x^{2} - 2 x + 9

f (x) = 320 x^{2} - 2560 x^{3}

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

What is the extreme f(x)=x^2sqrt(3-x) ?
The extreme f(x)=x^2sqrt(3-x) is Minimum(0,0),Maximum(12/5 ,(144sqrt(3))/(25sqrt(5))),Minimum(3,0)