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

The extreme f(x)=xsqrt(3x-x^2) is Minimum(0,0),Maximum(9/4 ,(27sqrt(3))/(16)),Minimum(3,0)

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

extreme points

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

Minimum (0, 0), Maximum (\frac{9}{4}, \frac{27 3}{16}), Minimum (3, 0)

Solution steps

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

Find intervals:

Increasing

: 0 < x < \frac{9}{4},

Decreasing

: \frac{9}{4} < x < 3

Plug

x = 0

into

x 3 x - x^{2} : 0

Minimum (0, 0)

Plug

x = \frac{9}{4}

into

x 3 x - x^{2} : \frac{27 3}{16}

Maximum (\frac{9}{4}, \frac{27 3}{16})

Plug

x = 3

into

x 3 x - x^{2} : 0

Minimum (3, 0)

Minimum (0, 0), Maximum (\frac{9}{4}, \frac{27 3}{16}), Minimum (3, 0)

y = cos (3 x) - 2

f (x) = - 3 x^{3} + 4 x^{2}, - 2 \leq x \leq 2

\frac{x ^{2} + 36}{x ^{2}}

f (x, y) = x e^{y} - x^{2} - e^{y}

f (x) = \frac{( 5 + x )}{( 3 - x )}

What is the extreme f(x)=xsqrt(3x-x^2) ?
The extreme f(x)=xsqrt(3x-x^2) is Minimum(0,0),Maximum(9/4 ,(27sqrt(3))/(16)),Minimum(3,0)