What is the extreme f(x)=x-\sqrt[3]{x},-1<= x<= 5 ?

The extreme f(x)=x-\sqrt[3]{x},-1<= x<= 5 is

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extreme f(x)=x-\sqrt[3]{x},-1<= x<= 5

extreme points

Solution steps

Find the critical points:

x = - 1, x = - \frac{1}{3 3}, x = 0, x = \frac{1}{3 3}, x = 5

Domain of

Find intervals:

Increasing

: - 1 < x < - \frac{1}{3 3},

Decreasing

: - \frac{1}{3 3} < x < 0,

Decreasing

: 0 < x < \frac{1}{3 3},

Increasing

: \frac{1}{3 3} < x < 5

Plug the extreme point

x = - 1

into

Minimum (- 1, 0)

Plug the extreme point

x = - \frac{1}{3 3}

into

Plug the extreme point

x = \frac{1}{3 3}

into

Plug the extreme point

x = 5

into

f (x) = 2 x + 7 y

f (x) = - 4 x^{2} - 2 y^{2} - 8 x + 12 y + 5

g (x) = \frac{x ^{3}}{( x + 1 )}

f (x) = x^{\frac{5}{4}} - 80 x^{\frac{1}{4}}

f (x) = 5 + 6 x - x^{2}, 0 \leq x \leq 4

What is the extreme f(x)=x-\sqrt[3]{x},-1<= x<= 5 ?
The extreme f(x)=x-\sqrt[3]{x},-1<= x<= 5 is