What is the extreme y=x^{4/5}(x-1) ?

The extreme y=x^{4/5}(x-1) is Maximum(0,0),Minimum(4/9 ,-(5*4^{4/5})/(9*9^{4/5)})

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extreme y=x^{4/5}(x-1)

extreme points

y = x^{\frac{4}{5}} (x - 1)

Maximum (0, 0), Minimum (\frac{4}{9}, - \frac{5 \cdot 4 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}})

Solution steps

f^{'} (x) = \frac{9 x - 4}{5 x ^{\frac{1}{5}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{4}{9},

Increasing

: \frac{4}{9} < x < \infty

Plug

x = 0

into

x^{\frac{4}{5}} (x - 1) : 0

Maximum (0, 0)

Plug

x = \frac{4}{9}

into

x^{\frac{4}{5}} (x - 1) : - \frac{5 \cdot 4 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}}

Minimum (\frac{4}{9}, - \frac{5 \cdot 4 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}})

Maximum (0, 0), Minimum (\frac{4}{9}, - \frac{5 \cdot 4 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}})

f (x) = \frac{( x ^{2} - 11 )}{x + 6}

\frac{64}{x ^{2} + 6 x - 7}

f (x) = 6 x e^{- x^{2}}

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

y = x^{3} - 12 x + 3

What is the extreme y=x^{4/5}(x-1) ?
The extreme y=x^{4/5}(x-1) is Maximum(0,0),Minimum(4/9 ,-(5*4^{4/5})/(9*9^{4/5)})