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

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

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

extreme points

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

Maximum (- 1, 3), Minimum (0, 0), Maximum (1, 3)

Solution steps

Find the critical points:

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

Domain of

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

Maximum (- 1, 3)

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = 1

into

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

Maximum (1, 3)

Maximum (- 1, 3), Minimum (0, 0), Maximum (1, 3)

(2 x - x^{2}) (2 y - y^{2})

sin (2 x)

f (x, y) = \frac{1}{3} x^{3} - 5 x^{2} - y^{2} - 3 y

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

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

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