Problem 13 - Calculus

Question

show that there is a number \(c,\) with \(0 \leq c \leq 1,\) such that \(f(c)=0\) $$f(x)=x-\cos x$$

Step-by-Step Solution

Answer

There exists a number \(c\) in \([0, 1]\) such that \(f(c) = 0\).

Step 1: Define the function

The function given is \( f(x) = x - \cos x \). First, we define the function and note that it is continuous because both \( x \) and \( \cos x \) are continuous everywhere.

Step 2: Evaluate the function at boundary points

Next, evaluate the function at the boundary points of the interval \([0, 1]\):\[ f(0) = 0 - \cos(0) = -1 \]\[ f(1) = 1 - \cos(1) \approx 1 - 0.54 = 0.46 \]So, \( f(0) < 0 \) and \( f(1) > 0 \).

Step 3: Apply the Intermediate Value Theorem

Since \( f(x) \) is continuous on the interval \([0, 1]\) and changes signs from negative to positive, we apply the Intermediate Value Theorem. This theorem states that if a continuous function changes signs over an interval, there must be at least one point \( c \) in the interval where the function equals zero.

Step 4: Conclude existence of \( c \)

According to the Intermediate Value Theorem, there is indeed a number \( c \), with \( 0 \leq c \leq 1 \), such that \( f(c) = 0 \). This means \( c - \cos c = 0 \), or equivalently, \( c = \cos c \).

Key Concepts

Continuous Function

A continuous function is a function where small changes in the input produce small changes in the output. This is crucial for the Intermediate Value Theorem, which relies on the function not having any "jumps" or "gaps" in its domain. In the example provided, the function is defined as \( f(x) = x - \cos x \). Both \( x \) and \( \cos x \) are continuous functions:

• The function \( x \) is simply a linear function of the form \( x = x \), and linear functions are continuous everywhere on their domain, which is typically all real numbers.
• The function \( \cos x \) is a trigonometric function, and it is continuous for all real numbers as well, seamlessly oscillating between -1 and 1.

When we combine these two functions, the resulting function \( f(x) = x - \cos x \) remains continuous everywhere, including within the interval [0, 1]. The continuity of \( f(x) \) is essential for using the Intermediate Value Theorem.

Boundary Points

Boundary points are the specific values that define the limits of an interval. In mathematical terms, these are often where we examine the behavior of a function to determine specific values or properties. In the exercise, we consider the boundary points of the interval \([0, 1]\). Here’s what we do with these points:

• Evaluate the function \( f(x) \) at \( x = 0 \): \[ f(0) = 0 - \cos(0) = -1 \]
• Evaluate the function \( f(x) \) at \( x = 1 \): \[ f(1) = 1 - \cos(1) \approx 0.46 \]

At \( x = 0 \), the function yields \( -1 \), which indicates that \( f(x) \) is negative. At \( x = 1 \), the function \( f(x) \) is positive or \( 0.46 \). These evaluations at boundary points help to show that there is a "change of sign" as we move from \( x = 0 \) to \( x = 1 \). As a result, the boundary points play a pivotal role in arguing for the existence of a root between them by using the Intermediate Value Theorem.

Change of Sign

The change of sign in a function is an important concept because it indicates that the function crosses the x-axis. This is relevant when working with the Intermediate Value Theorem, which uses the change of sign to confirm the existence of a root within an interval.
For a function to change sign over an interval, it means the function takes on both positive and negative values between the boundary points. In our problem, we have:

• At \( f(0) = -1 \), the function is negative.
• At \( f(1) = 0.46 \), the function is positive.

This sign change over the interval \([0, 1]\) confirms that somewhere between these two points, the function must cross the x-axis, i.e., \( f(x) = 0 \). According to the Intermediate Value Theorem, and given that \( f(x) \) is continuous, there exists at least one number \( c \) in \([0, 1]\) for which \( f(c) = 0 \). This is because a continuous function that goes from a negative value to a positive value must necessarily pass through zero.