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To determine if Rolle's theorem can be applied to the function \( f(x) = x^{2/3} - 6 \) on the interval \( [-27, 27] \), we need to check if the function meets the three criteria for Rolle's theorem:

1. **Continuity on the closed interval** \( [a, b] \).
2. **Differentiability on the open interval** \( (a, b) \).
3. **Equal values at the endpoints**, i.e., \( f(a) = f(b) \).

### Step 1: Check Continuity on \( [-27, 27] \)
The function \( f(x) = x^{2/3} - 6 \) is continuous for all real values of \( x \) because it is a transformation of the continuous function \( x^{2/3} \). Therefore, \( f(x) \) is continuous on the interval \( [-27, 27] \).

### Step 2: Check Differentiability on \( (-27, 27) \)
To check differentiability, we need the derivative of \( f(x) \):
\[
f(x) = x^{2/3} - 6
\]
\[
f'(x) = \frac{2}{3} x^{-1/3} = \frac{2}{3} \cdot \frac{1}{x^{1/3}}
\]
The derivative \( f'(x) = \frac{2}{3} \cdot \frac{1}{x^{1/3}} \) is undefined at \( x = 0 \) (since dividing by \( x^{1/3} \) when \( x = 0 \) is not possible). Thus, \( f(x) \) is **not differentiable** on the entire interval \( (-27, 27) \).

### Step 3: Check \( f(-27) = f(27) \)
Calculate the values at the endpoints:
\[
f(-27) = (-27)^{2/3} - 6 = 9 - 6 = 3
\]
\[
f(27) = (27)^{2/3} - 6 = 9 - 6 = 3
\]
Since \( f(-27) = f(27) = 3 \), the third condition is met.

### Conclusion
Rolle's theorem cannot be applied because \( f(x) \) is **not differentiable** on \( (-27, 27) \) due to the point at \( x = 0 \).

### Answer
The correct choices are:
- **No, because \( f \) is not differentiable on the open interval \( (a, b) \).**

For the value of \( c \):
- Since Rolle's theorem cannot be applied, enter **NA**.

Would you like further details or have any questions about this problem?

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1. What is Rolle's theorem and how does it apply to differentiable functions?
2. How do you determine the continuity of a function on a closed interval?
3. What makes a function not differentiable at certain points?
4. Why is \( x^{2/3} \) not differentiable at \( x = 0 \)?
5. Can there be cases where Rolle's theorem applies even if \( f'(c) = 0 \) at a point other than \( c = 0 \)?

**Tip:** For functions involving fractional exponents, check differentiability at zero carefully, as these functions often have issues at that point.

Determine whether Rolle's theorem can be applied to f(x) = x^(2/3) - 6 on the interval [-27, 27]. If Rolle's theorem can be applied, find all values of c in the open interval (a, b) such that f'(c) = 0. If Rolle's theorem cannot be applied, enter NA.

Explore whether Rolle's theorem can be applied to the function f(x) = x^(2/3) - 6 on the interval [-27, 27]. This problem requires an understanding of continuity, differentiability, and endpoint values. Detailed solution steps analyze the function's behavior and explain why the theorem is applicable or not.

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