The statement is true.

This conclusion comes from the application of the Mean Value Theorem. The theorem states that if a function $f$ is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists a point $c$ in $(a, b)$ where

$f'(c) = \frac{f(b) - f(a)}{b - a}.$

In this problem:

1. The function $f$ is given as differentiable, which implies it is also continuous.
2. We have $f(-6) = f(6)$, meaning that the values of $f$ at $x = -6$ and $x = 6$ are the same.

Since $f(6) = f(-6)$, we have

$f'(c) = \frac{f(6) - f(-6)}{6 - (-6)} = \frac{0}{12} = 0.$

Therefore, there must exist a point $c$ in the interval $(-6, 6)$ (where $|c| < 6$) such that $f'(c) = 0$, as required.

Would you like more details on this solution or have any questions?

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Here are 5 related questions for further exploration:

1. What are the conditions of the Mean Value Theorem and how can they be verified?
2. Can a function be continuous but not differentiable? Give examples.
3. How does the Mean Value Theorem apply to functions that are not differentiable at certain points within the interval?
4. If $f'(c) = 0$ at a certain point $c$, does it necessarily mean that $f$ has a local maximum or minimum at $c$?
5. How is Rolle’s Theorem a special case of the Mean Value Theorem?

Tip: The Mean Value Theorem is a powerful tool for proving the existence of points with specific derivative properties in an interval.