What is the extreme f(x)=x-\sqrt[3]{x},-1<= x<= 7 ?

The extreme f(x)=x-\sqrt[3]{x},-1<= x<= 7 is

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extreme f(x)=x-\sqrt[3]{x},-1<= x<= 7

extreme points

Solution steps

Find the critical points:

x = - 1, x = - \frac{1}{3 3}, x = 0, x = \frac{1}{3 3}, x = 7

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 < 7

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 = 7

into

f (x) = - x^{2} - y^{2} + 10 x + 10 y

2 cos (x) + cos^{2} (x)

x^{\frac{19}{9}} + x^{\frac{10}{9}}

- 10 x^{4} + 4 x^{2} - 6

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

What is the extreme f(x)=x-\sqrt[3]{x},-1<= x<= 7 ?
The extreme f(x)=x-\sqrt[3]{x},-1<= x<= 7 is