Question

problem 5 : determine if the mean value theorem can be applied to the following function on the given closed interval. if so, find all possible values of (c): (f(x) = \sin 2x) on (0, \pi)

Explanation:

🆕 New Concept Discovered: Mean Value Theorem
Guarantees a point where instantaneous rate equals average rate.

Step 1: Verify the hypotheses of the Mean Value Theorem

To apply the Mean Value Theorem (MVT) to \( f(x) = \sin 2x \) on the interval \( [0, \pi] \), two conditions must be met:

1. Continuity: The function \( f(x) \) must be continuous on the closed interval \( [0, \pi] \).
2. Differentiability: The function \( f(x) \) must be differentiable on the open interval \( (0, \pi) \).

Since \( f(x) = \sin 2x \) is a trigonometric sine function, it is continuous and differentiable everywhere on the real line. Therefore, both conditions are satisfied, and the MVT can be applied.

Step 2: Calculate the average rate of change

The MVT states that there exists at least one value \( c \) in the open interval \( (0, \pi) \) such that:

$$ f'(c) = \frac{f(\pi) - f(0)}{\pi - 0} $$

First, evaluate the function at the endpoints:

$$ f(0) = \sin(2 \cdot 0) = \sin 0 = 0 $$

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

Now, calculate the average rate of change:

$$ \frac{f(\pi) - f(0)}{\pi - 0} = \frac{0 - 0}{\pi} = 0 $$

Step 3: Find the derivative of the function

Find the derivative \( f'(x) \) using the chain rule:

$$ f'(x) = \frac{d}{dx}[\sin 2x] = 2\cos 2x $$

Step 4: Solve for \( c \)

Set the derivative equal to the average rate of change:

$$ f'(c) = 0 $$

$$ 2\cos 2c = 0 $$

$$ \cos 2c = 0 $$

We need to find all solutions for \( c \) in the open interval \( (0, \pi) \), which means \( 2c \) must lie in the interval \( (0, 2\pi) \).

The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):

$$ 2c = \frac{\pi}{2} \implies c = \frac{\pi}{4} $$

$$ 2c = \frac{3\pi}{2} \implies c = \frac{3\pi}{4} $$

Both values, \( c = \frac{\pi}{4} \) and \( c = \frac{3\pi}{4} \), lie within the open interval \( (0, \pi) \).

Answer:

The Mean Value Theorem can be applied because \( f(x) \) is continuous on \( [0, \pi] \) and differentiable on \( (0, \pi) \).

The values of \( c \) are:

$$ c = \frac{\pi}{4}, \quad c = \frac{3\pi}{4} $$