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

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

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

extreme points

Solution steps

Find the critical points:

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

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 < 3

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 = 3

into

x^{2} + x y + 3 x + 2 y + 5

f (x) = x^{3} + 2 x^{2} - 5 x - 6

f (x) = \frac{1}{x ^{2} - 9}

y = 2 x^{3} - 3 x + 5

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

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