What is the extreme f(t)=t-\sqrt[3]{t},-1<= t<= 4 ?

The extreme f(t)=t-\sqrt[3]{t},-1<= t<= 4 is

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extreme f(t)=t-\sqrt[3]{t},-1<= t<= 4

extreme points

f (t) = t - 3 t, - 1 \leq t \leq 4

Minimum (- 1, 0), Maximum (- \frac{1}{3 3}, - \frac{1}{3 3} + 3 \frac{1}{3 3}), Minimum (\frac{1}{3 3}, \frac{1}{3 3} - 3 \frac{1}{3 3}), Maximum (4, 4 - 34)

Solution steps

Find the critical points:

t = - 1, t = - \frac{1}{3 3}, t = 0, t = \frac{1}{3 3}, t = 4

Domain of

t - 3 t, - 1 \leq t \leq 4 : - 1 \leq t \leq 4

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 < 4

Plug the extreme point

x = - 1

into

t - 3 t \Rightarrow y = 0

Minimum (- 1, 0)

Plug the extreme point

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

into

t - 3 t \Rightarrow y = - \frac{1}{3 3} + 3 \frac{1}{3 3}

Maximum (- \frac{1}{3 3}, - \frac{1}{3 3} + 3 \frac{1}{3 3})

Plug the extreme point

x = \frac{1}{3 3}

into

t - 3 t \Rightarrow y = \frac{1}{3 3} - 3 \frac{1}{3 3}

Minimum (\frac{1}{3 3}, \frac{1}{3 3} - 3 \frac{1}{3 3})

Plug the extreme point

x = 4

into

t - 3 t \Rightarrow y = 4 - 34

Maximum (4, 4 - 34)

Minimum (- 1, 0), Maximum (- \frac{1}{3 3}, - \frac{1}{3 3} + 3 \frac{1}{3 3}), Minimum (\frac{1}{3 3}, \frac{1}{3 3} - 3 \frac{1}{3 3}), Maximum (4, 4 - 34)

e^{x} (x - 6)

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

e^{x} (x - 4)

f (x) = e^{x} (x - 4)

f (x) = 3 x + sin (6 x)

What is the extreme f(t)=t-\sqrt[3]{t},-1<= t<= 4 ?
The extreme f(t)=t-\sqrt[3]{t},-1<= t<= 4 is