What is the extreme f(x)=\sqrt[3]{x},27<= x<= 64 ?

The extreme f(x)=\sqrt[3]{x},27<= x<= 64 is Minimum(27,3),Maximum(64,4)

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extreme f(x)=\sqrt[3]{x},27<= x<= 64

extreme points

f (x) = 3 x, 27 \leq x \leq 64

Minimum (27, 3), Maximum (64, 4)

Solution steps

Find the critical points:

x = 27, x = 64

Domain of

3 x, 27 \leq x \leq 64 : 27 \leq x \leq 64

Find intervals:

Increasing

: 27 < x < 64

Plug the extreme point

x = 27

into

3 x \Rightarrow y = 3

Minimum (27, 3)

Plug the extreme point

x = 64

into

3 x \Rightarrow y = 4

Maximum (64, 4)

Minimum (27, 3), Maximum (64, 4)

f (x, y) = 2 x y^{2} - 2 y^{2} - 9 x^{2}

f (x) = 3 x (x - 1)

f (x) = x - 2 x, (0, 9)

f (x) = 2 x^{3} + 6 x^{2} - 48 x + 6

- 80 e^{x} + 2 x e^{x} + 2 x^{2} e^{x} - 1

What is the extreme f(x)=\sqrt[3]{x},27<= x<= 64 ?
The extreme f(x)=\sqrt[3]{x},27<= x<= 64 is Minimum(27,3),Maximum(64,4)