Problem 3 - Calculus Early Transcendentals

Question

Verify that the hypotheses of Rolle's Theorem are satisfied on the given interval, and find all values of \(c\) in that interval that satisfy the conclusion of the theorem. $$ f(x)=\cos x ;[\pi / 2,3 \pi / 2] $$

Step-by-Step Solution

Answer

Rolle's Theorem applies, and \( c = \pi \) satisfies it.

Step 1: Identify the Function and Interval

We have the function \( f(x) = \cos x \) and the interval \( [\frac{\pi}{2}, \frac{3\pi}{2}] \). First, we need to establish if this function is continuous and differentiable over this interval.

Step 2: Check Continuity and Differentiability

The cosine function \( f(x) = \cos x \) is continuous and differentiable everywhere, which includes the interval \( [\frac{\pi}{2}, \frac{3\pi}{2}] \). Thus, the first condition of Rolle's Theorem is satisfied.

Step 3: Evaluate Function at the Endpoints

Now, evaluate \( f(x) \) at the endpoints of the interval: \( f(\frac{\pi}{2}) = \cos\left(\frac{\pi}{2}\right) = 0 \) and \( f(\frac{3\pi}{2}) = \cos\left(\frac{3\pi}{2}\right) = 0 \). Since \( f(\frac{\pi}{2}) = f(\frac{3\pi}{2}) \), the second condition of Rolle's Theorem is satisfied.

Step 4: Apply Rolle's Theorem

Rolle's Theorem tells us there exists at least one \( c \) in \( (\frac{\pi}{2}, \frac{3\pi}{2}) \) such that the derivative \( f'(c) = 0 \).

Step 5: Find the Derivative

Compute the derivative of \( f(x) \): \( f'(x) = -\sin x \).

Step 6: Solve for \( c \) where \( f'(c) = 0 \)

Set \( f'(x) = -\sin x = 0 \). This implies \( \sin x = 0 \). The general solution is \( x = n\pi \) for integer \( n \). In the interval \( (\frac{\pi}{2}, \frac{3\pi}{2}) \), the solution is \( c = \pi \).

Step 7: Verify \( c \) is in Given Interval

Check that \( c = \pi \) is indeed in the open interval \( (\frac{\pi}{2}, \frac{3\pi}{2}) \), which it is.

Key Concepts

Understanding Continuity

In simple terms, continuity of a function means that you can draw the graph of the function without lifting your pencil off the paper.
For a function to be continuous on a given interval, every point within that interval must satisfy the condition:

• There are no jumps in the function values.
• There are no holes or breaks in the graph of the function.
• The function does not head off to infinity suddenly.

For the function in question, \(f(x) = \cos x\), it is crucial to know that trigonometric functions like cosine are continuous everywhere on their domain.
This includes the interval \([\frac{\pi}{2}, \frac{3\pi}{2}]\) specified in our problem. Therefore, \(f(x) = \cos x\) does not have any breaks, jumps, or infinite discontinuities within this interval, ensuring continuity.

Exploring Differentiability

Differentiability is a fancy word for saying that a function has a derivative at each point within an interval.
If a function is differentiable at a particular point, it means you can compute its slope or rate of change at that point.
It's important for a function to be both continuous and smooth (no sharp edges) to be differentiable.

• If a function is not continuous at a point, it can't possibly be differentiable there.
• However, a continuous function can still be non-differentiable if it has corners or cusps.

The function \(f(x) = \cos x\) is smooth without any sharp turns, making it differentiable everywhere on its domain.
The trigonometric function meets the requirements of both continuity and differentiability in the interval \([\frac{\pi}{2}, \frac{3\pi}{2}]\), satisfying the conditions of Rolle's Theorem.

Identifying Critical Points

Critical points are special points on the graph of the function where the function's derivative is zero or undefined. In other words, these points could signal the locations of peaks, troughs, or even points of inflection.

• To find critical points within the interval, you first compute the derivative of the function.
• Afterward, you set the derivative equal to zero and solve for \(x\).

For \(f(x) = \cos x\), the derivative is \(f'(x) = -\sin x\). To find the critical points:

• Set \(-\sin x = 0\), which results in \(\sin x = 0\).
• The general solution is \(x = n\pi\), where \(n\) is an integer.

Within our interval \((\frac{\pi}{2}, \frac{3\pi}{2})\), the only critical point occurs at \(x = \pi\).
This critical point is the value of \(c\) that satisfies Rolle's Theorem within the given interval.