Question

If f(x) = cos(x) on the interval (0,2$\pi$), what is the value of c in (0,2$\pi$) where f'(c) = 0, as guaranteed by Rolle's Theorem?

• A. c = $\pi$
• B. c = 3$\pi$/2
• C. c = 5$\pi$
• D. c = $\pi$/4

Answer 1

Answer: (A)

The value of c is $\pi$.

Answer 2

To find the value of $c$ as guaranteed by Rolle's Theorem, we first need to ensure that the conditions of Rolle's Theorem are met for the given function and interval.

Rolle's Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$:

1. $f(x)$ must be continuous on $[a, b]$.
2. $f(x)$ must be differentiable on $(a, b)$.
3. $f(a) = f(b)$.

If these three conditions are satisfied, then there exists at least one value $c$ in the open interval $(a, b)$ such that $f'(c) = 0$.

Applying Rolle's Theorem to $f(x) = \cos(x)$ on $(0, 2\pi)$:

The problem states the interval as $(0, 2\pi)$. For Rolle's Theorem, we consider the closed interval $[0, 2\pi]$.

1. Continuity: The function $f(x) = \cos(x)$ is continuous for all real numbers, so it is continuous on the closed interval $[0, 2\pi]$. (Condition 1 satisfied)

2. Differentiability: The function $f(x) = \cos(x)$ is differentiable for all real numbers. Its derivative is $f'(x) = -\sin(x)$, which exists for all $x$. Thus, $f(x)$ is differentiable on the open interval $(0, 2\pi)$. (Condition 2 satisfied)

3. Equality of function values at endpoints: $f(0) = \cos(0) = 1$ $f(2\pi) = \cos(2\pi) = 1$ Since $f(0) = f(2\pi)$, this condition is also satisfied.

Finding $c$ where $f'(c) = 0$:

Since all three conditions of Rolle's Theorem are met, there must exist at least one value $c \in (0, 2\pi)$ such that $f'(c) = 0$.

First, find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(\cos(x)) = -\sin(x)$

Now, set $f'(c) = 0$: $-\sin(c) = 0$ $\sin(c) = 0$

We need to find the values of $c$ in the interval $(0, 2\pi)$ for which $\sin(c) = 0$. The general solution for $\sin(c) = 0$ is $c = n\pi$, where $n$ is an integer.

Let's check values of $n$ to find $c$ within the interval $(0, 2\pi)$:

• If $n=0$, $c = 0\pi = 0$. This value is not in the open interval $(0, 2\pi)$.
• If $n=1$, $c = 1\pi = \pi$. This value is in the open interval $(0, 2\pi)$.
• If $n=2$, $c = 2\pi$. This value is not in the open interval $(0, 2\pi)$.
• For any other integer values of $n$, $c$ will be outside the interval $(0, 2\pi)$.

Therefore, the only value of $c$ in $(0, 2\pi)$ for which $f'(c) = 0$ is $c = \pi$.