What is the extreme f(x)=x^{2/3}(x-3) ?

The extreme f(x)=x^{2/3}(x-3) is Maximum(0,0),Minimum(6/5 ,-(9*6^{2/3})/(5*5^{2/3)})

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

extreme points

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

Maximum (0, 0), Minimum (\frac{6}{5}, - \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Solution steps

f^{'} (x) = \frac{5 x - 6}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{6}{5},

Increasing

: \frac{6}{5} < x < \infty

Plug

x = 0

into

x^{\frac{2}{3}} (x - 3) : 0

Maximum (0, 0)

Plug

x = \frac{6}{5}

into

x^{\frac{2}{3}} (x - 3) : - \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}}

Minimum (\frac{6}{5}, - \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Maximum (0, 0), Minimum (\frac{6}{5}, - \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

f (x) = 4 x^{3} - 24 x^{2} + 36 x

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

x^{3} - 12 x^{2} - 45 x + 8

g (t) = e^{4 t^{3}} x ln (3 t - 1)

f (x) = - 4 x^{3} + 6 x^{2} + 1, - 4 \leq x \leq 3

What is the extreme f(x)=x^{2/3}(x-3) ?
The extreme f(x)=x^{2/3}(x-3) is Maximum(0,0),Minimum(6/5 ,-(9*6^{2/3})/(5*5^{2/3)})