What is the extreme f(x)=x^{2/3},-1<= x<= 8 ?

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

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extreme f(x)=x^{2/3},-1<= x<= 8

extreme points

f (x) = x^{\frac{2}{3}}, - 1 \leq x \leq 8

Maximum (- 1, 1), Minimum (0, 0), Maximum (8, 4)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 8

Domain of

x^{\frac{2}{3}}, - 1 \leq x \leq 8 : - 1 \leq x \leq 8

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 8

Plug the extreme point

x = - 1

into

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

Maximum (- 1, 1)

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = 8

into

x^{\frac{2}{3}} \Rightarrow y = 4

Maximum (8, 4)

Maximum (- 1, 1), Minimum (0, 0), Maximum (8, 4)

2 x^{3} - 3 x^{2} - 72 x + 1

f (x) = \frac{x ^{5}}{5} - ln (x)

f (x, y) = 8 x^{4} - x^{2} + 3 y^{2}

f (x, y) = 3 x^{2} + y^{3} - 18 x y + 22

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