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To solve this problem using the Mean Value Theorem (MVT), let's go through the steps:

### Step 1: Mean Value Theorem (MVT) Conditions
The Mean Value Theorem states that if \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists at least one \( c \) in \((a, b)\) such that
\[
f'(c) = \frac{f(b) - f(a)}{b - a}.
\]
In this case:
- \( f(x) = 3\sqrt{x} - 5 \),
- The interval \([a, b] = [25, 81]\).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
1. \( f(25) = 3\sqrt{25} - 5 = 3 \cdot 5 - 5 = 15 - 5 = 10 \).
2. \( f(81) = 3\sqrt{81} - 5 = 3 \cdot 9 - 5 = 27 - 5 = 22 \).

So,
\[
f(25) = 10 \quad \text{and} \quad f(81) = 22.
\]

### Step 3: Calculate the Average Rate of Change
The average rate of change over \([25, 81]\) is:
\[
\frac{f(81) - f(25)}{81 - 25} = \frac{22 - 10}{81 - 25} = \frac{12}{56} = \frac{3}{14}.
\]

### Step 4: Find \( f'(x) \)
Now, we find the derivative \( f'(x) \) of \( f(x) = 3\sqrt{x} - 5 \):
1. Rewrite \( f(x) = 3x^{1/2} - 5 \).
2. Use the power rule: \( f'(x) = \frac{3}{2} x^{-1/2} = \frac{3}{2\sqrt{x}} \).

### Step 5: Set \( f'(c) = \frac{3}{14} \)
According to the MVT, we need \( f'(c) = \frac{3}{14} \):
\[
\frac{3}{2\sqrt{c}} = \frac{3}{14}.
\]
Solve for \( c \) by isolating \( \sqrt{c} \):
1. Cross-multiply to get: \( 3 \cdot 14 = 2\sqrt{c} \cdot 3 \),
2. Simplify: \( 42 = 6\sqrt{c} \),
3. Divide by 6: \( \sqrt{c} = 7 \),
4. Square both sides: \( c = 49 \).

### Answer
The value of \( c \) that satisfies the Mean Value Theorem on \([25, 81]\) is:
\[
c = 49.
\]

Would you like further clarification or have additional questions?

Here are some related questions that can deepen your understanding:

1. How would you apply the Mean Value Theorem if the function was not differentiable at some points within the interval?
2. What happens if the function is continuous but not differentiable over the interval?
3. Can the Mean Value Theorem be applied to intervals that include negative values for functions involving square roots?
4. How is the Mean Value Theorem used in real-life applications?
5. How does the Mean Value Theorem relate to Rolle's Theorem?

**Tip:** When dealing with square roots in MVT problems, ensure the interval includes values where the function is defined and differentiable, especially when the domain is restricted by the function itself.

Find the exact value(s) of all numbers c that satisfy the mean value theorem for f(x) = 3√x - 5 on [25, 81].

Find the exact value of c that satisfies the Mean Value Theorem for the function f(x) = 3√x - 5 on the interval [25, 81]. This calculus problem covers the Mean Value Theorem and differentiation techniques for functions with square roots. Suitable for advanced high school or college-level calculus students.

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