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The problem is asking to find the critical values of the function:

\[
f(x) = x^{1/5}(6 - 2x).
\]

### Solution:

1. **Find the derivative**:
   To find the critical values, we first need to find \( f'(x) \). We will use the product rule because \( f(x) \) is a product of \( x^{1/5} \) and \( (6 - 2x) \).

   Let:
   \[
   u = x^{1/5} \quad \text{and} \quad v = 6 - 2x.
   \]

   Then:
   \[
   u' = \frac{1}{5}x^{-4/5} \quad \text{and} \quad v' = -2.
   \]

   Using the product rule, \( (uv)' = u'v + uv' \):

   \[
   f'(x) = \frac{1}{5}x^{-4/5}(6 - 2x) + x^{1/5}(-2).
   \]

2. **Simplify the derivative**:
   \[
   f'(x) = \frac{1}{5}x^{-4/5}(6 - 2x) - 2x^{1/5}.
   \]

   To combine the terms, we factor out \( x^{-4/5} \):

   \[
   f'(x) = x^{-4/5} \left(\frac{1}{5}(6 - 2x) - 2x \cdot x^{1}\right).
   \]

   Simplifying further:

   \[
   f'(x) = x^{-4/5} \left( \frac{6 - 2x}{5} - 2x \right).
   \]

   Finding a common denominator:

   \[
   f'(x) = x^{-4/5} \left( \frac{6 - 2x - 10x}{5} \right).
   \]

   Which simplifies to:

   \[
   f'(x) = x^{-4/5} \left( \frac{6 - 12x}{5} \right).
   \]

   Thus:

   \[
   f'(x) = \frac{x^{-4/5}(6 - 12x)}{5}.
   \]

3. **Find critical values**:
   Critical values occur when \( f'(x) = 0 \) or when \( f'(x) \) is undefined.

   - Setting \( f'(x) = 0 \):

     \[
     \frac{x^{-4/5}(6 - 12x)}{5} = 0.
     \]

     This implies:

     \[
     6 - 12x = 0 \quad \Rightarrow \quad x = \frac{6}{12} = \frac{1}{2}.
     \]

   - Check when \( f'(x) \) is undefined:
     
     \( f'(x) \) is undefined when \( x^{-4/5} \) is undefined, i.e., when \( x = 0 \).

4. **Conclusion**:
   The two critical values of the function are:

   \[
   x = \frac{1}{2} \quad \text{and} \quad x = 0.
   \]

Would you like more details on any of these steps, or have any questions? 

Here are some additional questions related to this problem:

1. How do we determine if a critical point is a maximum, minimum, or saddle point?
2. What is the significance of the exponent \( \frac{1}{5} \) in \( x^{1/5} \) with respect to differentiability?
3. How does the product rule apply when one function has a negative exponent?
4. What is the behavior of the function \( f(x) \) near the critical points?
5. How would the critical values change if the function were \( x^{1/5}(6 + 2x) \)?

**Tip**: When finding critical points, always check if the function or its derivative is undefined at certain points, as these can also indicate critical values.

Find the two critical values of the function f(x) = x^(1/5)(6 - 2x).

Discover how to find the critical values of the function f(x) = x^(1/5)(6 - 2x). Learn step-by-step how to apply the product rule for differentiation and identify points where the function's derivative equals zero or is undefined. Suitable for undergraduate calculus students.

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