What is the extreme f(x)=\sqrt[3]{x},-1<= x<= 8 ?

The extreme f(x)=\sqrt[3]{x},-1<= x<= 8 is Minimum(-1,-1),Maximum(8,2)

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extreme f(x)=\sqrt[3]{x},-1<= x<= 8

extreme points

Minimum (- 1, - 1), Maximum (8, 2)

Solution steps

Find the critical points:

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

Domain of

Find intervals:

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < 8

Plug the extreme point

x = - 1

into

Minimum (- 1, - 1)

Plug the extreme point

x = 8

into

Maximum (8, 2)

Minimum (- 1, - 1), Maximum (8, 2)

F (x, y) = y^{3} + 3 x^{2} y - 6 x^{2} - 6 y^{2} + 2

f (x, y) = - x^{2} - 4 y^{2} + 6 x + 32 y - 76

f (x, y) = 3 x y^{2} - x^{3} - 15 x + 36 y + 9

f (x, y) = \frac{ln ( x )}{y - ∣ x ∣}

f (x) = 4 x^{\frac{2}{3}} - 5 x^{\frac{1}{3}}

What is the extreme f(x)=\sqrt[3]{x},-1<= x<= 8 ?
The extreme f(x)=\sqrt[3]{x},-1<= x<= 8 is Minimum(-1,-1),Maximum(8,2)