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

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

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

extreme points

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

Minimum (- 2, - 2^{\frac{2}{3}}), Maximum (- 1, 0), Minimum (- \frac{1}{2}, - (- \frac{1}{4})^{\frac{2}{3}}), Maximum (0, 0), Minimum (3, - 1 2^{\frac{2}{3}})

Solution steps

Find the critical points:

x = - 2, x = - 1, x = - \frac{1}{2}, x = 0, x = 3

Domain of

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

Find intervals:

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < - \frac{1}{2},

Increasing

: - \frac{1}{2} < x < 0,

Decreasing

: 0 < x < 3

Plug the extreme point

x = - 2

into

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

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

Plug the extreme point

x = - 1

into

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

Maximum (- 1, 0)

Plug the extreme point

x = - \frac{1}{2}

into

- (x^{2} + x)^{\frac{2}{3}} \Rightarrow y = - (- \frac{1}{4})^{\frac{2}{3}}

Minimum (- \frac{1}{2}, - (- \frac{1}{4})^{\frac{2}{3}})

Plug the extreme point

x = 0

into

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

Maximum (0, 0)

Plug the extreme point

x = 3

into

- (x^{2} + x)^{\frac{2}{3}} \Rightarrow y = - 1 2^{\frac{2}{3}}

Minimum (3, - 1 2^{\frac{2}{3}})

Minimum (- 2, - 2^{\frac{2}{3}}), Maximum (- 1, 0), Minimum (- \frac{1}{2}, - (- \frac{1}{4})^{\frac{2}{3}}), Maximum (0, 0), Minimum (3, - 1 2^{\frac{2}{3}})

52900 + 400 x + x^{2}

f (x) = - x - 4, - 4 \leq x \leq 1

f (x, y) = x^{2} - 9 x y - y^{2}

x θ

f (x) = 6 x^{3} - 18 x - 4

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