What is the extreme f(x)=x^{2/3}(x-6) ?

The extreme f(x)=x^{2/3}(x-6) is Maximum(0,0),Minimum(12/5 ,-(18*12^{2/3})/(5*5^{2/3)})

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extreme f(x)=x^{2/3}(x-6)

extreme points

f (x) = x^{\frac{2}{3}} (x - 6)

Maximum (0, 0), Minimum (\frac{12}{5}, - \frac{18 \cdot 1 2 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Solution steps

f^{'} (x) = \frac{5 x - 12}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{12}{5},

Increasing

: \frac{12}{5} < x < \infty

Plug

x = 0

into

x^{\frac{2}{3}} (x - 6) : 0

Maximum (0, 0)

Plug

x = \frac{12}{5}

into

x^{\frac{2}{3}} (x - 6) : - \frac{18 \cdot 1 2 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}}

Minimum (\frac{12}{5}, - \frac{18 \cdot 1 2 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Maximum (0, 0), Minimum (\frac{12}{5}, - \frac{18 \cdot 1 2 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

\frac{16 a ^{4} + 16 a}{a ^{2} + 1 - a}

f (x) = - 2 x^{2} - 5 x y - 6 y^{2} + 32 x + 63 y + 9

f (x) = 4 sin (x) - 3 cos (x)

y = (\frac{I x + 2 I - 1}{2 x + 3}) e^{2 x}

x + \frac{3}{2} x^{\frac{2}{3}}

What is the extreme f(x)=x^{2/3}(x-6) ?
The extreme f(x)=x^{2/3}(x-6) is Maximum(0,0),Minimum(12/5 ,-(18*12^{2/3})/(5*5^{2/3)})