What is the extreme f(x)=x^{1/3}(x+4),-27<= x<= 27 ?

The extreme f(x)=x^{1/3}(x+4),-27<= x<= 27 is Maximum(-27,69),Minimum(-1,-3),Maximum(27,93)

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extreme f(x)=x^{1/3}(x+4),-27<= x<= 27

extreme

Maximum (- 27, 69), Minimum (- 1, - 3), Maximum (27, 93)

Solution steps

Find the critical points:

x = - 27, x = - 1, x = 0, x = 27

Domain of

x^{\frac{1}{3}} (x + 4), - 27 \leq x \leq 27 : - 27 \leq x \leq 27

Find intervals:

Decreasing

: - 27 < x < - 1,

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < 27

Plug the extreme point

x = - 27

into

x^{\frac{1}{3}} (x + 4) \Rightarrow y = 69

Maximum (- 27, 69)

Plug the extreme point

x = - 1

into

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

Minimum (- 1, - 3)

Plug the extreme point

x = 27

into

x^{\frac{1}{3}} (x + 4) \Rightarrow y = 93

Maximum (27, 93)

Maximum (- 27, 69), Minimum (- 1, - 3), Maximum (27, 93)

What is the extreme f(x)=x^{1/3}(x+4),-27<= x<= 27 ?
The extreme f(x)=x^{1/3}(x+4),-27<= x<= 27 is Maximum(-27,69),Minimum(-1,-3),Maximum(27,93)