What is the extreme f(x)=3x^{2/3}-2x,-1<= x<= 1 ?

The extreme f(x)=3x^{2/3}-2x,-1<= x<= 1 is Maximum(-1,5),Minimum(0,0),Maximum(1,1)

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extreme f(x)=3x^{2/3}-2x,-1<= x<= 1

extreme points

f (x) = 3 x^{\frac{2}{3}} - 2 x, - 1 \leq x \leq 1

Maximum (- 1, 5), Minimum (0, 0), Maximum (1, 1)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1

Domain of

3 x^{\frac{2}{3}} - 2 x, - 1 \leq x \leq 1 : - 1 \leq x \leq 1

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 1

into

3 x^{\frac{2}{3}} - 2 x \Rightarrow y = 5

Maximum (- 1, 5)

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = 1

into

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

Maximum (1, 1)

Maximum (- 1, 5), Minimum (0, 0), Maximum (1, 1)

f (x, y) = 1 - x^{2} - y^{2}

f (x) = x^{3} - 9 x^{2} + 24 x - 7

f (x) = \frac{1}{x ^{3} - 6 x ^{2} + 1}

f (x, y) = 1 - x^{2} - 1 - y^{2}

f (x) = x^{3} - 3 x^{2} + 3 x + 1

What is the extreme f(x)=3x^{2/3}-2x,-1<= x<= 1 ?
The extreme f(x)=3x^{2/3}-2x,-1<= x<= 1 is Maximum(-1,5),Minimum(0,0),Maximum(1,1)