- Verify continuity and differentiability of \(f(x) = \sin(8x)\) on the interval \([\frac{\pi}{8}, \frac{\pi}{4}]\).
- Confirm \(f(\frac{\pi}{8}) = f(\frac{\pi}{4}) = 0\), satisfying the conditions for Rolle's Theorem.
- Compute the derivative \(f'(x) = 8\cos(8x)\) and set it to zero to find critical points.
- Identify the value \(c = \frac{3\pi}{16}\) as the solution within the interval \(\boxed{\frac{3\pi}{16}}\).
Explanation
- Verifying Conditions of Rolle's Theorem
Rolle's Theorem states that for a function \(f(x)\) to satisfy the theorem on a closed interval \([a, b]\), three conditions must be met:
- \(f(x)\) must be continuous on the closed interval \([a, b]\).
- \(f(x)\) must be differentiable on the open interval \((a, b)\).
- \(f(a)\) must equal \(f(b)\).
In this case, the function is \(f(x) = \sin(8x)\) on the interval \([\frac{\pi}{8}, \frac{\pi}{4}]\).
- Since \(\sin(8x)\) is a trigonometric function, it is continuous and differentiable for all real numbers. Thus, it is continuous on \([\frac{\pi}{8}, \frac{\pi}{4}]\) and differentiable on \((\frac{\pi}{8}, \frac{\pi}{4})\).
- Checking the function values at the endpoints:
\(f\left(\frac{\pi}{8}\right) = \sin(8 \cdot \frac{\pi}{8}) = \sin(\pi) = 0\)
\(f\left(\frac{\pi}{4}\right) = \sin(8 \cdot \frac{\pi}{4}) = \sin(2\pi) = 0\)
Since \(f(\frac{\pi}{8}) = f(\frac{\pi}{4}) = 0\), all conditions of Rolle's Theorem are satisfied.
- Finding the Derivative and Critical Points
Since Rolle's Theorem applies, there exists at least one number \(c\) in the open interval \((\frac{\pi}{8}, \frac{\pi}{4})\) such that \(f'(c) = 0\).
First, find the derivative of \(f(x)\):
\(f'(x) = \frac{d}{dx}(\sin(8x)) = 8\cos(8x)\)
Now, set the derivative equal to zero to find the critical point(s):
\(8\cos(8c) = 0 \implies \cos(8c) = 0\)
The general solution for \(\cos(\theta) = 0\) is \(\theta = \frac{\pi}{2} + k\pi\) for any integer \(k\).
Substituting \(8c = \theta\):
\(8c = \frac{\pi}{2} + k\pi \implies c = \frac{\pi}{16} + \frac{k\pi}{8}\)
- Determining the Value of c
We need to find the values of \(k\) such that \(c\) lies in the interval \((\frac{\pi}{8}, \frac{\pi}{4})\).
Note that \(\frac{\pi}{8} = \frac{2\pi}{16}\) and \(\frac{\pi}{4} = \frac{4\pi}{16}\).
We require \(\frac{2\pi}{16} < \frac{\pi}{16} + \frac{2k\pi}{16} < \frac{4\pi}{16}\).
Subtracting \(\frac{\pi}{16}\) from all sides:
\(\frac{\pi}{16} < \frac{2k\pi}{16} < \frac{3\pi}{16}\)
Dividing by \(\frac{\pi}{16}\):
\(1 < 2k < 3\)
This gives \(k = 1\).
Substituting \(k=1\) into the expression for \(c\):
\(c = \frac{\pi}{16} + \frac{1\pi}{8} = \frac{\pi}{16} + \frac{2\pi}{16} = \frac{3\pi}{16}\).
Since \(\frac{3\pi}{16}\) is between \(\frac{2\pi}{16}\) (\(\frac{\pi}{8}\)) and \(\frac{4\pi}{16}\) (\(\frac{\pi}{4}\)), this value of \(c\) satisfies the requirements.
- Final Answer
Rolle's Theorem applies to \(f(x) = \sin(8x)\) on the interval \([\frac{\pi}{8}, \frac{\pi}{4}]\) because the function is continuous, differentiable, and \(f(\frac{\pi}{8}) = f(\frac{\pi}{4}) = 0\). The point \(c\) that satisfies Rolle's Theorem is \(\boxed{\frac{3\pi}{16}}\).
Examples
Rolle's Theorem is fundamental in physics and engineering, particularly in motion analysis. For instance, if an object starts at a specific position at time \(t=0\) and returns to that same position at time \(t=T\), Rolle's Theorem guarantees that there was at least one moment in time within \((0, T)\) where the object's instantaneous velocity was zero. This helps engineers identify peak positions or turning points in mechanical systems or electrical oscillators, ensuring they understand the behavior of the system between known states.