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

The extreme f(x)=x(1-x)^{2/5} is Maximum(5/7 ,(5*2^{2/5})/(7*7^{2/5)}),Minimum(1,0)

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

extreme points

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

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

Solution steps

f^{'} (x) = \frac{- 7 x + 5}{5 ( 1 - x ) ^{\frac{3}{5}}}

Find intervals:

Increasing

: - \infty < x < \frac{5}{7},

Decreasing

: \frac{5}{7} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = \frac{5}{7}

into

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

Maximum (\frac{5}{7}, \frac{5 \cdot 2 ^{\frac{2}{5}}}{7 \cdot 7 ^{\frac{2}{5}}})

Plug

x = 1

into

x (1 - x)^{\frac{2}{5}} : 0

Minimum (1, 0)

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

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

\frac{6}{- 3 x + 2}

ln (x^{2} - 4)

x^{2} y + y^{2} - 4 y

x^{2} + y^{2} + 4 x y

What is the extreme f(x)=x(1-x)^{2/5} ?
The extreme f(x)=x(1-x)^{2/5} is Maximum(5/7 ,(5*2^{2/5})/(7*7^{2/5)}),Minimum(1,0)