What is the extreme f(x)=(x^2-x)^{2/3} ?

The extreme f(x)=(x^2-x)^{2/3} is Minimum(0,0),Maximum(1/2 ,(-1/4)^{2/3}),Minimum(1,0)

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extreme f(x)=(x^2-x)^{2/3}

extreme points

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

Minimum (0, 0), Maximum (\frac{1}{2}, (- \frac{1}{4})^{\frac{2}{3}}), Minimum (1, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

(x^{2} - x)^{\frac{2}{3}} : 0

Minimum (0, 0)

Plug

x = \frac{1}{2}

into

(x^{2} - x)^{\frac{2}{3}} : (- \frac{1}{4})^{\frac{2}{3}}

Maximum (\frac{1}{2}, (- \frac{1}{4})^{\frac{2}{3}})

Plug

x = 1

into

(x^{2} - x)^{\frac{2}{3}} : 0

Minimum (1, 0)

Minimum (0, 0), Maximum (\frac{1}{2}, (- \frac{1}{4})^{\frac{2}{3}}), Minimum (1, 0)

f (x) = x + 1

f (x) = x e^{- 5 x^{2}}

f (x) = x^{3} - 12 x^{2} + 1

f (x) = x^{2} + 4 x + 6

1 - x^{\frac{2}{3}}

What is the extreme f(x)=(x^2-x)^{2/3} ?
The extreme f(x)=(x^2-x)^{2/3} is Minimum(0,0),Maximum(1/2 ,(-1/4)^{2/3}),Minimum(1,0)