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

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

Decreasing

: - 3 < x < 0,

Increasing

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

Decreasing

: \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}}

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

Plug the extreme point

x = 0

into

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

Minimum (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}}}

Maximum (\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

Minimum (3, 0)

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

x^{\frac{3}{5}}

f (x) = \frac{8.7 x + 0.2}{x ^{2} - 4.5 x + 7.5}

f (t) = \frac{e ^{t}}{t ^{2} + 1}, (0, 3)

f (x) = 5 x^{3} - 5 x^{2} - 5 x + 7

x^{2} - 2 x + 1

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