To determine if the Mean Value Theorem (MVT) applies to the function $f(x) = \frac{x^2 - 1}{x^2 - 25}$ on the interval $[0, 10]$, we need to check two main conditions of the MVT:

1. Continuity on the closed interval $[0, 10]$.
2. Differentiability on the open interval $(0, 10)$.

Step 1: Check Continuity

The function $f(x) = \frac{x^2 - 1}{x^2 - 25}$ will be continuous on $[0, 10]$ if the denominator $x^2 - 25$ is not zero within this interval. Observing that $x^2 - 25 = 0$ when $x = \pm 5$, there is a discontinuity at $x = 5$. Therefore, the function is not continuous on $[0, 10]$ because $x = 5$ lies within the interval.

Conclusion

Since $f(x)$ is not continuous on the interval $[0, 10]$, the Mean Value Theorem does not apply.

Thus, we should enter:

$c = \text{DNE}$

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Additional Related Questions

1. What are the implications of a discontinuity in terms of the Mean Value Theorem?
2. How would the answer change if the interval were modified to $[0, 4]$?
3. Can the MVT apply if the interval contains a point where the function is undefined?
4. What is an example of a function and interval where MVT definitely applies?
5. How does differentiability differ from continuity in the context of the MVT?

Tip: The Mean Value Theorem requires both continuity and differentiability on the given interval; missing either condition invalidates its application.