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The problem asks us to apply the Mean Value Theorem (MVT) to the function \( f(x) = \sin(\pi x) \) over the interval \( [0, 2] \), and to find all points \( c \in (0, 2) \) such that:

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

### Step 1: Recall 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 a point \( c \in (a, b) \) such that:
\[
f'(c) = \frac{f(b) - f(a)}{b - a}.
\]
In this case, we are looking for \( f'(c) \) for the interval \( [0, 2] \).

### Step 2: Calculate \( f(2) \) and \( f(0) \)
Given \( f(x) = \sin(\pi x) \), let's compute:
- \( f(2) = \sin(\pi \cdot 2) = \sin(2\pi) = 0 \),
- \( f(0) = \sin(\pi \cdot 0) = \sin(0) = 0 \).

Thus,
\[
f(2) - f(0) = 0 - 0 = 0.
\]

### Step 3: Compute the derivative \( f'(x) \)
Next, we differentiate \( f(x) = \sin(\pi x) \):
\[
f'(x) = \pi \cos(\pi x).
\]

### Step 4: Set up the Mean Value Theorem equation
The equation from the MVT becomes:
\[
0 = f'(c)(2 - 0),
\]
which simplifies to:
\[
f'(c) = 0.
\]

### Step 5: Solve for \( c \)
Now, we solve \( f'(c) = 0 \):
\[
\pi \cos(\pi c) = 0.
\]
This implies:
\[
\cos(\pi c) = 0.
\]
The cosine function equals zero when \( \pi c = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \). However, since \( c \) must be in \( (0, 2) \), the only solution is:
\[
\pi c = \frac{\pi}{2} \implies c = \frac{1}{2}.
\]

### Final Answer:
The point \( c \) such that \( f(2) - f(0) = f'(c)(2 - 0) \) is \( c = \frac{1}{2} \).

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

Here are 5 related questions you might find interesting:
1. How does the Mean Value Theorem differ from Rolle's Theorem?
2. What are the conditions for a function to apply the Mean Value Theorem?
3. Can you apply the Mean Value Theorem on intervals where the function is not differentiable?
4. How would the solution change if the function was \( f(x) = \cos(\pi x) \)?
5. What is the geometric interpretation of the Mean Value Theorem?

**Tip:** The Mean Value Theorem can help you understand the behavior of a function's slope over an interval, and it guarantees at least one point where the instantaneous rate of change equals the average rate of change.

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).

This problem demonstrates how to apply the Mean Value Theorem to the trigonometric function f(x) = sin(πx) on the interval [0, 2]. The solution includes differentiating the function and finding the critical point c such that the theorem holds.

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