What is the extreme f(x)=\sqrt[3]{x},-1<= x<= 1 ?

The extreme f(x)=\sqrt[3]{x},-1<= x<= 1 is Minimum(-1,-1),Maximum(1,1)

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extreme f(x)=\sqrt[3]{x},-1<= x<= 1

extreme

Minimum (- 1, - 1), Maximum (1, 1)

Solution steps

Find the critical points:

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

Domain of

3 x, - 1 \leq x \leq 1 : - 1 \leq x \leq 1

Find intervals:

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 1

into

3 x \Rightarrow y = - 1

Minimum (- 1, - 1)

Plug the extreme point

x = 1

into

3 x \Rightarrow y = 1

Maximum (1, 1)

Minimum (- 1, - 1), Maximum (1, 1)

What is the extreme f(x)=\sqrt[3]{x},-1<= x<= 1 ?
The extreme f(x)=\sqrt[3]{x},-1<= x<= 1 is Minimum(-1,-1),Maximum(1,1)