To solve for the absolute maximum and minimum values of the function $f(x) = x^{1/3}$ on the interval $[1, 4]$, we will follow these steps:

1. Find the critical points:

To find the critical points, we first compute the derivative of the function $f(x) = x^{1/3}$.

$f'(x) = \frac{d}{dx} \left( x^{1/3} \right) = \frac{1}{3}x^{-2/3} = \frac{1}{3} \cdot \frac{1}{x^{2/3}}$

This derivative exists for all $x > 0$, and it never equals zero (since the numerator is a constant). Therefore, there are no critical points within the interval.

2. Evaluate the function at the endpoints:

Since there are no critical points, the absolute maximum and minimum must occur at the endpoints of the interval. We now evaluate the function $f(x) = x^{1/3}$ at $x = 1$ and $x = 4$:

• At $x = 1$: $f(1) = 1^{1/3} = 1$

• At $x = 4$: $f(4) = 4^{1/3} = \sqrt[3]{4}$

3. Conclusion:

• The absolute minimum value occurs at $x = 1$, and the value is $1$.
• The absolute maximum value occurs at $x = 4$, and the value is $\sqrt[3]{4}$.

Final Answer:

• Absolute maximum value: $\sqrt[3]{4}$
• Absolute minimum value: $1$

Would you like any further details or clarification?

Here are 5 follow-up questions based on this topic:

1. How do you find critical points of a function?
2. Why do we evaluate the function at the endpoints when finding absolute extrema?
3. What does it mean if a function's derivative is never zero within an interval?
4. How would the process change if the function had critical points within the interval?
5. Can you always assume the absolute extremum occurs at endpoints if there are no critical points?

Tip: When dealing with power functions like $x^{1/n}$, the derivative rule $x^{k}$ applies as usual: $\frac{d}{dx} x^{k} = kx^{k-1}$.