What is the extreme f(x)=8x^{1/3}-x^{4/3},-8<= x<= 8 ?

The extreme f(x)=8x^{1/3}-x^{4/3},-8<= x<= 8 is Minimum(-8,-32),Maximum(2,6*2^{1/3}),Minimum(8,0)

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extreme f(x)=8x^{1/3}-x^{4/3},-8<= x<= 8

extreme points

f (x) = 8 x^{\frac{1}{3}} - x^{\frac{4}{3}}, - 8 \leq x \leq 8

Minimum (- 8, - 32), Maximum (2, 6 \cdot 2^{\frac{1}{3}}), Minimum (8, 0)

Solution steps

Find the critical points:

x = - 8, x = 0, x = 2, x = 8

Domain of

8 x^{\frac{1}{3}} - x^{\frac{4}{3}}, - 8 \leq x \leq 8 : - 8 \leq x \leq 8

Find intervals:

Increasing

: - 8 < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 8

Plug the extreme point

x = - 8

into

8 x^{\frac{1}{3}} - x^{\frac{4}{3}} \Rightarrow y = - 32

Minimum (- 8, - 32)

Plug the extreme point

x = 2

into

8 x^{\frac{1}{3}} - x^{\frac{4}{3}} \Rightarrow y = 6 2^{\frac{1}{3}}

Maximum (2, 6 \cdot 2^{\frac{1}{3}})

Plug the extreme point

x = 8

into

8 x^{\frac{1}{3}} - x^{\frac{4}{3}} \Rightarrow y = 0

Minimum (8, 0)

Minimum (- 8, - 32), Maximum (2, 6 \cdot 2^{\frac{1}{3}}), Minimum (8, 0)

What is the extreme f(x)=8x^{1/3}-x^{4/3},-8<= x<= 8 ?
The extreme f(x)=8x^{1/3}-x^{4/3},-8<= x<= 8 is Minimum(-8,-32),Maximum(2,6*2^{1/3}),Minimum(8,0)