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We are tasked with using the Mean Value Theorem (MVT) to find all points \( 0 < c < 2 \) such that:

\[
f(2) - f(0) = f'(c)(2 - 0)
\]

where \( f(x) = \sin(\pi x) \).

### Step 1: Apply the Mean Value Theorem
The Mean Value Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists some \( c \in (a, b) \) such that:

\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]

In this case:
- \( f(x) = \sin(\pi x) \)
- The interval is \( [0, 2] \)
- So, \( a = 0 \) and \( b = 2 \).

### Step 2: Calculate \( f(2) \) and \( f(0) \)
We need to calculate \( f(2) \) and \( f(0) \):

- \( f(2) = \sin(2\pi) = 0 \)
- \( f(0) = \sin(0) = 0 \)

Thus, the left-hand side of the equation becomes:

\[
f(2) - f(0) = 0 - 0 = 0
\]

### Step 3: Differentiate \( f(x) = \sin(\pi x) \)
Next, we find the derivative of \( f(x) \):

\[
f'(x) = \frac{d}{dx}[\sin(\pi x)] = \pi \cos(\pi x)
\]

### Step 4: Set up the equation for \( c \)
Using the Mean Value Theorem, we know there exists \( c \in (0, 2) \) such that:

\[
f'(c)(2 - 0) = f(2) - f(0)
\]

Substitute what we know:

\[
\pi \cos(\pi c) \cdot 2 = 0
\]

This simplifies to:

\[
\pi \cos(\pi c) = 0
\]

### Step 5: Solve for \( c \)
Since \( \pi \neq 0 \), we must have:

\[
\cos(\pi c) = 0
\]

The cosine function equals zero at \( \pi c = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \). Solving for \( c \), we find:

\[
c = \frac{1}{2}, \frac{3}{2}, \dots
\]

However, since we are only interested in \( c \in (0, 2) \), the only valid solution is:

\[
c = \frac{1}{2}
\]

### Conclusion
The point \( c = \frac{1}{2} \) satisfies the conditions of the Mean Value Theorem for the given function on the interval \( (0, 2) \).

Would you like further details or have any questions?

Here are five related questions for deeper understanding:
1. How does the Mean Value Theorem relate to the concept of differentiability?
2. Can the Mean Value Theorem be applied to non-differentiable functions?
3. What is the geometrical interpretation of the Mean Value Theorem?
4. Why does \( \cos(\pi c) = 0 \) imply specific values for \( c \)?
5. How does the result change if we use a different interval, such as \( [1, 3] \)?

Tip: The Mean Value Theorem is a generalization of Rolle's Theorem, which applies when \( f(a) = f(b) \).

Let f(x) = sin(πx). Use the Mean Value Theorem to find all points 0 < c < 2 such that f(2) - f(0) = f'(c)(2 - 0).

Solve for c using the Mean Value Theorem for the function f(x) = sin(πx) on the interval (0, 2). This problem involves calculating derivatives and applying key trigonometric identities.

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