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 $f(a) = f(b)$.