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 points

f (x) = x - 3 x, - 1 \leq x \leq 6

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)

f (x, y) = 5 x^{2} - 4 x y + 2 y^{2} - 36 x

f (x, y) = 37 - x^{2} - y^{2}

f (x) = 0.09 x^{2} + 16 x + 350, 0 < x < 150

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

f (x, y) = x^{3} - y^{2} - 6 x y - 4

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