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

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

Minimum (0, 0), Maximum (\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:

Decreasing

: - \infty < x < 0,

Increasing

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

Decreasing

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

Plug

x = 0

into

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

Minimum (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}}}

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

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

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

\frac{x ^{2} - 7 x + 26}{x - 5}

f (x) = 3 x^{3} e^{- x}, - 1 \leq x \leq 5

(x + 8)^{\frac{2}{3}}

f (x) = x^{2} - y^{2} - 7 x - 4 y

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