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

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

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

extreme points

Minimum (- 27, - 3), Maximum (27, 3)

Solution steps

Find the critical points:

x = - 27, x = 0, x = 27

Domain of

Find intervals:

Increasing

: - 27 < x < 0,

Increasing

: 0 < x < 27

Plug the extreme point

x = - 27

into

Minimum (- 27, - 3)

Plug the extreme point

x = 27

into

Maximum (27, 3)

Minimum (- 27, - 3), Maximum (27, 3)

0.001 x^{2} + 7 + \frac{128}{x}

f (t) = \frac{1}{9} [\frac{t ^{3}}{3} + \frac{3 t ^{2}}{2} - 28 t]

F (X, Y) = X \cdot Y

u (x, y) = (2 - x) (y - 1) + 2 (x - 2)^{2} + 2 (y - 1)^{2} - 3

g (x) = x^{6} - 2 x^{5} + 8 x^{4}

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