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

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

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extreme f(x)=x-\sqrt[3]{x},-1<= x<= 4

extreme points

Solution steps

Find the critical points:

x = - 1, x = - \frac{1}{3 3}, x = 0, x = \frac{1}{3 3}, x = 4

Domain of

Find intervals:

Increasing

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

Decreasing

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

Decreasing

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

Increasing

: \frac{1}{3 3} < x < 4

Plug the extreme point

x = - 1

into

Minimum (- 1, 0)

Plug the extreme point

x = - \frac{1}{3 3}

into

Plug the extreme point

x = \frac{1}{3 3}

into

Plug the extreme point

x = 4

into

2 x^{3} - 27 x^{2} + 84 x, 1 \leq x \leq 8

f (x, y) = 1 0^{- 4} y - 2 \cdot 1 0^{- 8} y^{2}

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

f (x) = x^{2} lo g_{10} (2 x)

f (x) = - [x + 2]

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