What is the extreme f(x)=-x^{2/3}(x-1) ?

The extreme f(x)=-x^{2/3}(x-1) is Minimum(0,0),Maximum(2/5 ,(3*2^{2/3})/(5*5^{2/3)})

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extreme f(x)=-x^{2/3}(x-1)

extreme points

f (x) = - x^{\frac{2}{3}} (x - 1)

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

Solution steps

f^{'} (x) = - \frac{5 x - 2}{3 x ^{\frac{1}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{2}{5},

Decreasing

: \frac{2}{5} < x < \infty

Plug

x = 0

into

- x^{\frac{2}{3}} (x - 1) : 0

Minimum (0, 0)

Plug

x = \frac{2}{5}

into

- x^{\frac{2}{3}} (x - 1) : \frac{3 \cdot 2 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}}

Maximum (\frac{2}{5}, \frac{3 \cdot 2 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

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

f (x) = x^{3} (x - 6)^{4}

f (x) = \frac{2}{3} x^{3} - 2 x^{2} + \frac{44}{3}

f (x) = 2 x^{3} - 24 x^{2} + 72 x + 7

y = x^{2} lo g_{6} (x)

f (x) = x x^{2} + 9

What is the extreme f(x)=-x^{2/3}(x-1) ?
The extreme f(x)=-x^{2/3}(x-1) is Minimum(0,0),Maximum(2/5 ,(3*2^{2/3})/(5*5^{2/3)})