What is the extreme f(x)=x^{2/3}(x-3),-3<= x<= 3 ?

The extreme f(x)=x^{2/3}(x-3),-3<= x<= 3 is

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extreme f(x)=x^{2/3}(x-3),-3<= x<= 3

extreme points

f (x) = x^{\frac{2}{3}} (x - 3), - 3 \leq x \leq 3

Minimum (- 3, - 6 (- 3)^{\frac{2}{3}}), Maximum (0, 0), Minimum (\frac{6}{5}, - \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}}), Maximum (3, 0)

Solution steps

Find the critical points:

x = - 3, x = 0, x = \frac{6}{5}, x = 3

Domain of

x^{\frac{2}{3}} (x - 3), - 3 \leq x \leq 3 : - 3 \leq x \leq 3

Find intervals:

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < \frac{6}{5},

Increasing

: \frac{6}{5} < x < 3

Plug the extreme point

x = - 3

into

x^{\frac{2}{3}} (x - 3) \Rightarrow y = - 6 (- 3)^{\frac{2}{3}}

Minimum (- 3, - 6 (- 3)^{\frac{2}{3}})

Plug the extreme point

x = 0

into

x^{\frac{2}{3}} (x - 3) \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = \frac{6}{5}

into

x^{\frac{2}{3}} (x - 3) \Rightarrow y = - \frac{9 6 ^{\frac{2}{3}}}{5 5 ^{\frac{2}{3}}}

Minimum (\frac{6}{5}, - \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Plug the extreme point

x = 3

into

x^{\frac{2}{3}} (x - 3) \Rightarrow y = 0

Maximum (3, 0)

Minimum (- 3, - 6 (- 3)^{\frac{2}{3}}), Maximum (0, 0), Minimum (\frac{6}{5}, - \frac{9 \cdot 6 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}}), Maximum (3, 0)

f (x) = 2 x^{3} - 27 x^{2} + 84 x + 11

f (x) = x 2 x - x^{2}

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

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

f (x) = 10 x^{2} y - 5 x^{2} - 4 y^{2} - x^{4} - 2 y^{4}

What is the extreme f(x)=x^{2/3}(x-3),-3<= x<= 3 ?
The extreme f(x)=x^{2/3}(x-3),-3<= x<= 3 is