What is the extreme f(x)=3xsqrt(4-3x) ?

The extreme f(x)=3xsqrt(4-3x) is Maximum(8/9 ,(16)/(3sqrt(3))),Minimum(4/3 ,0)

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extreme f(x)=3xsqrt(4-3x)

extreme

Maximum (\frac{8}{9}, \frac{16}{3 3}), Minimum (\frac{4}{3}, 0)

Solution steps

f^{'} (x) = \frac{3 ( - 9 x + 8 )}{2 4 - 3 x}

Find intervals:

Increasing

: - \infty < x < \frac{8}{9},

Decreasing

: \frac{8}{9} < x < \frac{4}{3}

Plug

x = \frac{8}{9}

into

3 x 4 - 3 x : \frac{16}{3 3}

Maximum (\frac{8}{9}, \frac{16}{3 3})

Plug

x = \frac{4}{3}

into

3 x 4 - 3 x : 0

Minimum (\frac{4}{3}, 0)

Maximum (\frac{8}{9}, \frac{16}{3 3}), Minimum (\frac{4}{3}, 0)

What is the extreme f(x)=3xsqrt(4-3x) ?
The extreme f(x)=3xsqrt(4-3x) is Maximum(8/9 ,(16)/(3sqrt(3))),Minimum(4/3 ,0)