What is the extreme f(x)=x^{2/5}(x-8) ?

The extreme f(x)=x^{2/5}(x-8) is Maximum(0,0),Minimum(16/7 ,-(40*16^{2/5})/(7*7^{2/5)})

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extreme f(x)=x^{2/5}(x-8)

extreme points

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

Maximum (0, 0), Minimum (\frac{16}{7}, - \frac{40 \cdot 1 6 ^{\frac{2}{5}}}{7 \cdot 7 ^{\frac{2}{5}}})

Solution steps

f^{'} (x) = \frac{7 x - 16}{5 x ^{\frac{3}{5}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{16}{7},

Increasing

: \frac{16}{7} < x < \infty

Plug

x = 0

into

x^{\frac{2}{5}} (x - 8) : 0

Maximum (0, 0)

Plug

x = \frac{16}{7}

into

x^{\frac{2}{5}} (x - 8) : - \frac{40 \cdot 1 6 ^{\frac{2}{5}}}{7 \cdot 7 ^{\frac{2}{5}}}

Minimum (\frac{16}{7}, - \frac{40 \cdot 1 6 ^{\frac{2}{5}}}{7 \cdot 7 ^{\frac{2}{5}}})

Maximum (0, 0), Minimum (\frac{16}{7}, - \frac{40 \cdot 1 6 ^{\frac{2}{5}}}{7 \cdot 7 ^{\frac{2}{5}}})

What is the extreme f(x)=x^{2/5}(x-8) ?
The extreme f(x)=x^{2/5}(x-8) is Maximum(0,0),Minimum(16/7 ,-(40*16^{2/5})/(7*7^{2/5)})