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

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

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extreme y=xsqrt(4-x)

extreme points

y = x 4 - x

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < \frac{8}{3},

Decreasing

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

Plug

x = \frac{8}{3}

into

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

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

Plug

x = 4

into

x 4 - x : 0

Minimum (4, 0)

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

f (x) = (3 x + 1)^{5} (2 x + 3)^{2} (2 - x)^{3}

y = 2 x^{3} - 12 x^{2} - 270 x + 2

y = 0.0004 x^{2} - 0.0118 x + 2.6821

x (\frac{8}{5} x^{3} - 2 x^{2})

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

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