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

The extreme f(x)=x^{4/3},-1<= x<= 8 is Maximum(-1,1),Minimum(0,0),Maximum(8,16)

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

extreme points

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

Maximum (- 1, 1), Minimum (0, 0), Maximum (8, 16)

Solution steps

Find the critical points:

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

Domain of

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

Maximum (- 1, 1)

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = 8

into

x^{\frac{4}{3}} \Rightarrow y = 16

Maximum (8, 16)

Maximum (- 1, 1), Minimum (0, 0), Maximum (8, 16)

f (x, y) = (x + y - 1) \cdot x y

f (x) = x^{3} + x^{2} - 5 x - 4

f (x) = x 36 - x^{2}, - 6 \leq x \leq 6

f (x) = 5 x y

f (x, y) = x^{3} + y^{3} - 192 x - 300 y + 2

What is the extreme f(x)=x^{4/3},-1<= x<= 8 ?
The extreme f(x)=x^{4/3},-1<= x<= 8 is Maximum(-1,1),Minimum(0,0),Maximum(8,16)