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

The extreme f(x)=x^{1/5}(x+1) is Minimum(-1/6 ,-5/(6*6^{1/5)})

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

extreme points

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

Minimum (- \frac{1}{6}, - \frac{5}{6 \cdot 6 ^{\frac{1}{5}}})

Solution steps

f^{'} (x) = \frac{6 x + 1}{5 x ^{\frac{4}{5}}}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{6},

Increasing

: - \frac{1}{6} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{1}{6}

into

x^{\frac{1}{5}} (x + 1) : - \frac{5}{6 \cdot 6 ^{\frac{1}{5}}}

Minimum (- \frac{1}{6}, - \frac{5}{6 \cdot 6 ^{\frac{1}{5}}})

P (x, y) = 0.4 x + 0.6 y

f (x) = (1 + x)^{2} (6 - x)

f (x) = - 3 x^{2} - x + 2, - 4 \leq x \leq 4

\frac{7 + 2 x}{x}

f (x) = \frac{( e ^{x} )}{7 + e ^{x}}

What is the extreme f(x)=x^{1/5}(x+1) ?
The extreme f(x)=x^{1/5}(x+1) is Minimum(-1/6 ,-5/(6*6^{1/5)})