What is the extreme f(x)=x^{4/5}(x-2) ?

The extreme f(x)=x^{4/5}(x-2) is Maximum(0,0),Minimum(8/9 ,-(10*8^{4/5})/(9*9^{4/5)})

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extreme f(x)=x^{4/5}(x-2)

extreme points

f (x) = x^{\frac{4}{5}} (x - 2)

Maximum (0, 0), Minimum (\frac{8}{9}, - \frac{10 \cdot 8 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}})

Solution steps

f^{'} (x) = \frac{9 x - 8}{5 x ^{\frac{1}{5}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{8}{9},

Increasing

: \frac{8}{9} < x < \infty

Plug

x = 0

into

x^{\frac{4}{5}} (x - 2) : 0

Maximum (0, 0)

Plug

x = \frac{8}{9}

into

x^{\frac{4}{5}} (x - 2) : - \frac{10 \cdot 8 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}}

Minimum (\frac{8}{9}, - \frac{10 \cdot 8 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}})

Maximum (0, 0), Minimum (\frac{8}{9}, - \frac{10 \cdot 8 ^{\frac{4}{5}}}{9 \cdot 9 ^{\frac{4}{5}}})

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

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

f (x) = 8 x - 16 cos (x)

x^{2} + x y + y^{2} + 3 y

f (x, y) = x^{2} + x y + y^{2} - 9

What is the extreme f(x)=x^{4/5}(x-2) ?
The extreme f(x)=x^{4/5}(x-2) is Maximum(0,0),Minimum(8/9 ,-(10*8^{4/5})/(9*9^{4/5)})