What is the extreme f(x)=\sqrt[3]{x}(x-1) ?

The extreme f(x)=\sqrt[3]{x}(x-1) is Minimum(1/4 ,-3/(4\sqrt[3]{4)})

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extreme f(x)=\sqrt[3]{x}(x-1)

extreme points

f (x) = 3 x (x - 1)

Minimum (\frac{1}{4}, - \frac{3}{4 3 4})

Solution steps

Find the critical points:

x = 0, x = \frac{1}{4}

Domain of

3 x (x - 1) : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{1}{4},

Increasing

: \frac{1}{4} < x < \infty

Plug the extreme point

x = \frac{1}{4}

into

3 x (x - 1) \Rightarrow y = - \frac{3}{4 3 4}

Minimum (\frac{1}{4}, - \frac{3}{4 3 4})

f (x) = x - 2 x, (0, 9)

f (x) = 2 x^{3} + 6 x^{2} - 48 x + 6

- 80 e^{x} + 2 x e^{x} + 2 x^{2} e^{x} - 1

f (x) = x^{\frac{2}{3}} (\frac{5}{2} + x)

f (x) = \frac{x}{x ^{2} - x + 16}, 0 \leq x \leq 18

What is the extreme f(x)=\sqrt[3]{x}(x-1) ?
The extreme f(x)=\sqrt[3]{x}(x-1) is Minimum(1/4 ,-3/(4\sqrt[3]{4)})