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

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

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

extreme

Minimum (- 2, - 4 (- 2)^{\frac{2}{3}}), Maximum (0, 0), Minimum (\frac{4}{5}, - \frac{6 \cdot 4 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}}), Maximum (2, 0)

Solution steps

Find the critical points:

x = - 2, x = 0, x = \frac{4}{5}, x = 2

Domain of

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

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \frac{4}{5},

Increasing

: \frac{4}{5} < x < 2

Plug the extreme point

x = - 2

into

x^{\frac{2}{3}} (x - 2) \Rightarrow y = - 4 (- 2)^{\frac{2}{3}}

Minimum (- 2, - 4 (- 2)^{\frac{2}{3}})

Plug the extreme point

x = 0

into

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

Maximum (0, 0)

Plug the extreme point

x = \frac{4}{5}

into

x^{\frac{2}{3}} (x - 2) \Rightarrow y = - \frac{6 4 ^{\frac{2}{3}}}{5 5 ^{\frac{2}{3}}}

Minimum (\frac{4}{5}, - \frac{6 \cdot 4 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Plug the extreme point

x = 2

into

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

Maximum (2, 0)

Minimum (- 2, - 4 (- 2)^{\frac{2}{3}}), Maximum (0, 0), Minimum (\frac{4}{5}, - \frac{6 \cdot 4 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}}), Maximum (2, 0)

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