Find the absolute maximum and absolute minimum values of f on the given interval.
f(t) = t − t^1/3
,    [−1, 5]

To find the absolute maximum and minimum values of the function \( f(t) = t - t^{1/3} \) on the interval \([-1, 5]\), we need to follow these steps:

1. **Find the derivative**: We start by finding the derivative of \( f(t) \):
   \[
   f'(t) = 1 - \frac{1}{3} t^{-2/3} = 1 - \frac{1}{3 \sqrt[3]{t^2}}
   \]

2. **Set the derivative to zero**: We set \( f'(t) = 0 \) to find critical points:
   \[
   1 - \frac{1}{3 \sqrt[3]{t^2}} = 0
   \]
   Solving for \( t \):
   \[
   \frac{1}{3 \sqrt[3]{t^2}} = 1 \quad \Rightarrow \quad 3 = \sqrt[3]{t^2} \quad \Rightarrow \quad t^2 = 27 \quad \Rightarrow \quad t = \pm 3\sqrt{3}
   \]
   Since \( 3\sqrt{3} \approx 5.196 \) is outside the interval, we only consider \( t = -3\sqrt{3} \) which is also not within the interval.

3. **Evaluate the endpoints**: Now, we evaluate \( f(t) \) at the endpoints of the interval:
   - For \( t = -1 \):
     \[
     f(-1) = -1 - (-1)^{1/3} = -1 + 1 = 0
     \]
   - For \( t = 5 \):
     \[
     f(5) = 5 - 5^{1/3} = 5 - \sqrt[3]{5} \approx 5 - 1.710 \approx 3.290
     \]

4. **Compare values**: The values at the endpoints are:
   - \( f(-1) = 0 \)
   - \( f(5) \approx 3.290 \)

Since there are no critical points in the interval \([-1, 5]\), the absolute maximum and minimum values are determined by the endpoints.

5. **Conclusion**:
   - **Absolute maximum**: \( f(5) \approx 3.290 \)
   - **Absolute minimum**: \( f(-1) = 0 \)

### Summary
- Absolute maximum: \( \approx 3.290 \) at \( t = 5 \)
- Absolute minimum: \( 0 \) at \( t = -1 \)

If you have any questions or need further details, feel free to ask!

### Additional Questions:
1. How do you find critical points of a function?
2. What is the significance of the derivative in determining maximum and minimum values?
3. Can you explain how to evaluate a function at specific points?
4. What are the differences between local and absolute extrema?
5. How do you confirm if a point is a maximum or minimum using the second derivative test?

### Tip
Always check the endpoints of your interval and any critical points found within that interval when determining absolute maxima and minima.

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = t − t^1/3, [−1, 5]

Explore the process of finding absolute maximum and minimum values of the function f(t) = t - t^(1/3) on the interval [-1, 5]. This problem includes critical point analysis and endpoint evaluations, suitable for Grades 11-12.

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