Question

Use the Intermediate Value Theorem and Rolle's Theorem to prove that the equation has exactly one real Solution. $$ 2 x-2-\cos x=0 $$

Short answer

The equation \(2x-2-\cos x = 0\) has exactly one real root on the interval [0,2].

Step-by-step solution

Set the Function

Define the function of the equation: \(f(x) = 2x - 2 - \cos x\).

Find Values of f(x) that Exceed 0

Notice that \(f(0) = -3\) and \(f(2) = 1\). This ensures that the function \(f(x)\) moves from negative to positive values, which surpasses the x-axis at some point (due to the disjointed continuity of \(f(x)\)). Thus, a value c exists in the interval (0, 2) such that \(f(c) = 0\). This observation is based on the Intermediate Value Theorem.

Prove the Uniqueness of the Solution

To show that the value of c is unique, assume that there is another root d where \(d \neq c\) and d is also in the interval (0,2). Apply Rolle's theorem on the interval [c,d] or [d,c] (depending on which is greater). At minimum one point exists in this interval where the derivative of \(f(x)\) equals to 0. However, the derivative of \(f(x)\) is \(f'(x) = 2 + \sin x\), and this is greater than 0 for all x, which contradicts the Rolle's theorem. Thus, our assumption of a second root d is wrong.

Conclude the Proof

Therefore, the equation \(2x-2-\cos x = 0\) has exactly one real root on the interval [0,2].

Key concepts

Rolle's Theorem

When it comes to understanding the underpinnings of calculus, Rolle's Theorem stands out as a fundamental concept. At its core, it provides a kind of guarantee about the behavior of differentiable functions. To invoke this theorem, you need a function that is continuous on the closed interval \[a, b\] and differentiable on the open interval \(a, b\), with equal values at the endpoints, meaning that \(f(a) = f(b)\).

Rolle's Theorem asserts that there is at least one point \(c\) in the open interval \(a, b\) where the derivative \(f'(c)\) is zero. This is where the function's slope is flat, or in other words, the tangent to the curve at \(c\) is horizontal. In the context of real solutions to equations, this theorem helps us determine if there is a unique solution in a given interval by ensuring that if multiple solutions existed, there would have to be a point where the derivative equals zero. If such a point doesn't exist, the solution must be unique.

Real Solutions of Equations

The quest for real solutions of equations is akin to finding the 'x-marks-the-spot' on a treasure map; these solutions are the x-values where the function curve intersects the x-axis. In mathematical terms, we say that \(f(x) = 0\) for real solutions.

When assessing equations for real solutions, one valuable tool is the Intermediate Value Theorem. It posits that for any function continuous on a closed interval, if the function at one endpoint is negative and at the other endpoint is positive, then the function must have at least one root in that interval. However, this doesn't specify the number of real solutions. Here, concepts like Rolle's Theorem can further aid in deducing the uniqueness of the solution, contributing to a deeper understanding of how functions behave within the interval and ensuring that we can precisely pinpoint the x that marks the spot.

Derivative

One cannot talk about calculus without the concept of a derivative showing up. It's the mathematical equivalent of a detective following the clues. What a derivative does is it measures the rate at which a function's output changes as its input changes. Loosely speaking, it calculates the slope of the function at any given point.

The derivative is often denoted as \(f'(x)\) and is defined as the limit of the average rate of change of the function over a small interval as the interval shrinks to zero. This can be visualized as the slope of the tangent line to the function's graph at a particular point \(x\). In our original problem, the derivative of \(f(x) = 2x - 2 - \text{cos} x\) is \(f'(x) = 2 + \text{sin} x\). By showing this derivative is always positive, we're essentially proving that the function constantly increases, and therefore a repeated crossing of the x-axis (multiple real solutions) is impossible, affirming the uniqueness of the solution.

Continuity in Calculus

Continuity in Calculus is a promise of smooth sailing through the values of a function. A continuous function is one that has no breaks, jumps, or holes in its graph. Put simply, you could draw its curve without lifting your pen off the paper. For a function to be continuous at a point, the function's value at that point must be defined, and it must be equal to the limit as you approach that point from both sides.

In the realm of the Intermediate Value Theorem, which relies on continuity, this seamless passage is crucial. For example, in our exercise, the function \(f(x) = 2x - 2 - \text{cos} x\) is continuous over the real numbers because both components, the polynomial and the cosine function, are continuous. Continuity guarantees that our function will pass through every value between \(f(a)\) and \(f(b)\) when \(a\) and \(b\) are points in the domain of \(f\). This unbroken journey underpins the conviction that a real solution exists within the interval we are exploring and sets the stage for Rolle's Theorem to shine a light on the uniqueness of that solution.