What is the extreme f(x)=\sqrt[3]{x},-8<= x<= 8 ?

The extreme f(x)=\sqrt[3]{x},-8<= x<= 8 is Minimum(-8,-2),Maximum(8,2)

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extreme f(x)=\sqrt[3]{x},-8<= x<= 8

extreme points

f (x) = 3 x, - 8 \leq x \leq 8

Minimum (- 8, - 2), Maximum (8, 2)

Solution steps

Find the critical points:

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

Domain of

3 x, - 8 \leq x \leq 8 : - 8 \leq x \leq 8

Find intervals:

Increasing

: - 8 < x < 0,

Increasing

: 0 < x < 8

Plug the extreme point

x = - 8

into

3 x \Rightarrow y = - 2

Minimum (- 8, - 2)

Plug the extreme point

x = 8

into

3 x \Rightarrow y = 2

Maximum (8, 2)

Minimum (- 8, - 2), Maximum (8, 2)

f (x) = x^{\frac{2}{3}} - 7

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

f (x, y) = 2 x^{2} + 4 y^{2} + \frac{1}{2}

f (x) = \frac{x ^{2} + 2}{x + 1}

f (x) = 2 x - 11 ln (8 x), 2 \leq x \leq 9

What is the extreme f(x)=\sqrt[3]{x},-8<= x<= 8 ?
The extreme f(x)=\sqrt[3]{x},-8<= x<= 8 is Minimum(-8,-2),Maximum(8,2)