To find the absolute maximum and minimum values of the function \( f(t) = t - \sqrt[3]{t} \) on the interval \([-1, 5]\), we need to evaluate the function at the critical points and the endpoints of the interval. Critical points are found by setting the derivative of the function to zero and solving for \( t \).
To find the critical points of the function \( f(t) = t - \sqrt[3]{t} \), we first need to find its derivative and set it to zero.
\[
f(t) = t - t^{1/3}
\]
\[
f'(t) = 1 - \frac{1}{3}t^{-2/3}
\]
\[
f'(t) = 1 - \frac{1}{3}t^{-2/3} = 1 - \frac{1}{3} \cdot \frac{1}{t^{2/3}}
\]
Set the derivative equal to zero to find the critical points:
\[
1 - \frac{1}{3t^{2/3}} = 0
\]
\[
1 = \frac{1}{3t^{2/3}}
\]
\[
3t^{2/3} = 1
\]
\[
t^{2/3} = \frac{1}{3}
\]
\[
t = \left(\frac{1}{3}\right)^{3/2}
\]
\[
t = \frac{1}{\sqrt{27}} = \frac{1}{3\sqrt{3}} = \frac{\sqrt{3}}{9}
\]
Next, we evaluate the function \( f(t) \) at the critical point and at the endpoints of the interval \([-1, 5]\).
- At \( t = -1 \):
\[
f(-1) = -1 - \sqrt[3]{-1} = -1 + 1 = 0
\]
- At \( t = 5 \):
\[
f(5) = 5 - \sqrt[3]{5}
\]
- At \( t = \frac{\sqrt{3}}{9} \):
\[
f\left(\frac{\sqrt{3}}{9}\right) = \frac{\sqrt{3}}{9} - \sqrt[3]{\frac{\sqrt{3}}{9}}
\]
We need to simplify and compare the values obtained:
- \( f(-1) = 0 \)
- \( f(5) = 5 - \sqrt[3]{5} \)
- \( f\left(\frac{\sqrt{3}}{9}\right) \)
To find \( f\left(\frac{\sqrt{3}}{9}\right) \):
\[
f\left(\frac{\sqrt{3}}{9}\right) = \frac{\sqrt{3}}{9} - \left(\frac{\sqrt{3}}{9}\right)^{1/3}
\]
\[
\left(\frac{\sqrt{3}}{9}\right)^{1/3} = \left(\frac{\sqrt{3}}{9}\right)^{1/3} = \left(\frac{\sqrt{3}}{3^2}\right)^{1/3} = \frac{(\sqrt{3})^{1/3}}{3^{2/3}}
\]
\[
= \frac{3^{1/6}}{3^{2/3}} = 3^{1/6 - 4/6} = 3^{-1/2} = \frac{1}{\sqrt{3}}
\]
\[
f\left(\frac{\sqrt{3}}{9}\right) = \frac{\sqrt{3}}{9} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{9} - \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{9} - \frac{3\sqrt{3}}{9} = -\frac{2\sqrt{3}}{9}
\]
Comparing the values:
- \( f(-1) = 0 \)
- \( f(5) = 5 - \sqrt[3]{5} \)
- \( f\left(\frac{\sqrt{3}}{9}\right) = -\frac{2\sqrt{3}}{9} \)
The absolute maximum value is \( f(5) = 5 - \sqrt[3]{5} \).
The absolute minimum value is \( f\left(\frac{\sqrt{3}}{9}\right) = -\frac{2\sqrt{3}}{9} \).
\[
\boxed{\text{Absolute maximum: } 5 - \sqrt[3]{5}}
\]
\[
\boxed{\text{Absolute minimum: } -\frac{2\sqrt{3}}{9}}
\]
![Find the absolute maximum and absolute minimum values of the function
f(t)=t-t^(1/3)
on the interval [-1,5]. Enter only the y-value as your answer.
Absolute maximum:
Absolute minimum:](https://thumbnail.justsolvely.com/seoQuestionThumb/99dab329-d080-478b-b7cb-f7026d5909ba.webp)
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