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To solve for the **absolute maximum** and **absolute minimum** of the function \( f(t) = t - \sqrt[3]{t} \) on the interval \([-1, 6]\), we will follow these steps:

### Step 1: Find the derivative of the function
First, we need to find the critical points by taking the derivative of \( f(t) = t - \sqrt[3]{t} \).

\[
f(t) = t - t^{1/3}
\]

The derivative is:

\[
f'(t) = 1 - \frac{1}{3}t^{-2/3}
\]

Simplifying:

\[
f'(t) = 1 - \frac{1}{3\sqrt[3]{t^2}}
\]

### Step 2: Set the derivative equal to 0 to find critical points
We solve \( f'(t) = 0 \):

\[
1 - \frac{1}{3\sqrt[3]{t^2}} = 0
\]

\[
\frac{1}{3\sqrt[3]{t^2}} = 1
\]

\[
\sqrt[3]{t^2} = \frac{1}{3}
\]

Cubing both sides:

\[
t^2 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}
\]

\[
t = \pm \frac{1}{\sqrt[3]{27}} = \pm \frac{1}{3}
\]

So, the critical points are \( t = \frac{1}{3} \) and \( t = -\frac{1}{3} \).

### Step 3: Evaluate the function at critical points and endpoints
We now evaluate \( f(t) \) at the critical points and the endpoints \( t = -1 \) and \( t = 6 \).

- \( f(-1) = -1 - \sqrt[3]{-1} = -1 + 1 = 0 \)
- \( f(6) = 6 - \sqrt[3]{6} \approx 6 - 1.817 \approx 4.183 \)
- \( f\left(\frac{1}{3}\right) = \frac{1}{3} - \sqrt[3]{\frac{1}{3}} \approx 0.333 - 0.693 \approx -0.36 \)
- \( f\left(-\frac{1}{3}\right) = -\frac{1}{3} - \sqrt[3]{-\frac{1}{3}} = -\frac{1}{3} + 0.693 \approx 0.36 \)

### Step 4: Determine absolute minimum and maximum
- The minimum value is \( f\left(\frac{1}{3}\right) \approx -0.36 \).
- The maximum value is \( f(6) \approx 4.183 \).

Thus:
- The **absolute minimum** is approximately \(-0.36\).
- The **absolute maximum** is approximately \(4.183\).

Let me know if you need further clarification or additional details!

### Follow-up questions:
1. What are the exact critical points of a function with a cubic root?
2. Why are the endpoints also evaluated for absolute extrema?
3. How does the cubic root affect the function's behavior near zero?
4. Can the second derivative test confirm the nature of the critical points here?
5. How would you apply the same method to a different interval, e.g., \([0, 5]\)?

### Tip:
Always check both the critical points and the boundary points of the interval when looking for absolute extrema!

Find the absolute maximum and absolute minimum values of f(t) = t - ³√t on the interval [-1, 6]

Learn how to find the absolute maximum and minimum values of the function f(t) = t - ³√t on the interval [-1, 6]. This problem involves calculating derivatives, finding critical points, and evaluating at boundary points. Suitable for calculus students.

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