What is the extreme f(x)=x^{1/5}(x-2) ?

The extreme f(x)=x^{1/5}(x-2) is Minimum(1/3 ,-5/(3*3^{1/5)})

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extreme f(x)=x^{1/5}(x-2)

extreme points

f (x) = x^{\frac{1}{5}} (x - 2)

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

Solution steps

f^{'} (x) = \frac{2 ( 3 x - 1 )}{5 x ^{\frac{4}{5}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{1}{3},

Increasing

: \frac{1}{3} < x < \infty

Plug

x = \frac{1}{3}

into

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

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

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

f (x) = \frac{( 7 ln ( x ) + 4 )}{x}

f (x) = \frac{24}{x} + 4 π r^{2}

f (x) = x^{3} - 36 x^{2} + 324 x - 3280.5

h (v, t) = v t - 5 t^{2}

What is the extreme f(x)=x^{1/5}(x-2) ?
The extreme f(x)=x^{1/5}(x-2) is Minimum(1/3 ,-5/(3*3^{1/5)})