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

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

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

extreme points

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

Maximum (- 1, - 2)

Plug

x = 1

into

Minimum (1, 2)

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

1 - \frac{1}{1 - x}

3 x + 5

f (x) = x^{2} - 5

x^{2} + 1

f (x) = 4 x + 5

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