What is the extreme t-\sqrt[3]{t},-1<= t<= 3 ?

The extreme t-\sqrt[3]{t},-1<= t<= 3 is

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extreme t-\sqrt[3]{t},-1<= t<= 3

extreme points

Solution steps

Find the critical points:

t = - 1, t = - \frac{1}{3 3}, t = 0, t = \frac{1}{3 3}, t = 3

Domain of

Find intervals:

Increasing

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

Decreasing

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

Decreasing

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

Increasing

: \frac{1}{3 3} < t < 3

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 = 3

into

540 x - 5 x^{3}

f (x) = 232 x - 5000 - x^{2}

f (x) = 5 x^{\frac{4}{5}} - x, 0 \leq x \leq 3025

\frac{e ^{2 x}}{x}

6 x^{2} - 12

What is the extreme t-\sqrt[3]{t},-1<= t<= 3 ?
The extreme t-\sqrt[3]{t},-1<= t<= 3 is