What is the extreme f(x)=x-\sqrt[3]{x},-1<= x<= 6 ?

The extreme f(x)=x-\sqrt[3]{x},-1<= x<= 6 is

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extreme f(x)=x-\sqrt[3]{x},-1<= x<= 6

extreme

Minimum (- 1, 0), Maximum (- \frac{1}{3 3}, - \frac{1}{3 3} + 3 \frac{1}{3 3}), Minimum (\frac{1}{3 3}, \frac{1}{3 3} - 3 \frac{1}{3 3}), Maximum (6, 6 - 36)

Solution steps

Find the critical points:

x = - 1, x = - \frac{1}{3 3}, x = 0, x = \frac{1}{3 3}, x = 6

Domain of

x - 3 x, - 1 \leq x \leq 6 : - 1 \leq x \leq 6

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 < 6

Plug the extreme point

x = - 1

into

x - 3 x \Rightarrow y = 0

Minimum (- 1, 0)

Plug the extreme point

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

into

x - 3 x \Rightarrow y = - \frac{1}{3 3} + 3 \frac{1}{3 3}

Maximum (- \frac{1}{3 3}, - \frac{1}{3 3} + 3 \frac{1}{3 3})

Plug the extreme point

x = \frac{1}{3 3}

into

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

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

Plug the extreme point

x = 6

into

x - 3 x \Rightarrow y = 6 - 36

Maximum (6, 6 - 36)

Minimum (- 1, 0), Maximum (- \frac{1}{3 3}, - \frac{1}{3 3} + 3 \frac{1}{3 3}), Minimum (\frac{1}{3 3}, \frac{1}{3 3} - 3 \frac{1}{3 3}), Maximum (6, 6 - 36)

What is the extreme f(x)=x-\sqrt[3]{x},-1<= x<= 6 ?
The extreme f(x)=x-\sqrt[3]{x},-1<= x<= 6 is