Problem 68 - Calculus of a Single Variable

Question

Use the Intermediate Value Theorem and Rolle’s Theorem to prove that the equation has exactly one real solution. \(2 x-2-\cos x=0\)

Step-by-Step Solution

Answer

The function \(f(x) = 2x - 2 - \cos x = 0\) has exactly one real solution.

Step 1: Apply the Intermediate Value Theorem

Begin with the function \(f(x) = 2x-2-\cos x\). Evaluate this function at two different points, say \(x = 0\) and \(x = 1\), respectively. For \(x = 0\), \(f(0) = 2(0) - 2 - \cos(0) = -3\). For \(x = 1\), \(f(1) = 2(1) - 2 - \cos(1) = 2 - 2 - 0.5403 = -0.5403\). Here \(f(0) < 0\) and \(f(1) > 0\). According to the Intermediate Value Theorem, at least one solution must exist between these two points since the function is continuous.

Step 2: Apply Rolle’s Theorem

Now, to prove uniqueness of the solution, use Rolle's Theorem. First ensure that \(f(x)\) fulfils the conditions of Rolle's Theorem, it must be continuous on the closed interval [0,1] and differentiable in the open interval (0,1). Since \(f(x)\) contains simple polynomial and cosine function, it is both continuous and differentiable on the set of real numbers. Now calculate its derivative \(f'(x) = 2 + \sin x\). There is only one solution \(x = \pi\) at which \(f'(x) = 0\). Hence, by Rolle’s Theorem, the function \(f(x) = 2x - 2 - \cos x\) has a unique solution.

Key Concepts

Rolle's Theorem

Rolle's Theorem is a principle that offers key insights into the behavior of differentiable functions. It states that if a function is both continuous on a closed interval \[a, b\] and differentiable on the open interval \(a, b\), with equal values at the endpoints \( f(a) = f(b) \), there exists at least one point \(c\) in the open interval \(a, b\) where the derivative \(f'(c)\) is zero. This critical point indicates that there is a horizontal tangent line to the function at \(c\), which suggests a point of 'flatness' where the function's slope is zero.

In the exercise, Rolle's Theorem helps us establish the uniqueness of the solution to the equation \(2x-2-\cos x=0\) by showing that there exists only one point where the derivative of the function is zero, thus confirming the existence of exactly one real solution.

Differentiability

Differentiability refers to the ability of a function to have a derivative at each point within its domain. When a function is differentiable, the tangent at each point is well-defined and the function has no sharp turns or cusps. It's a stronger condition than continuity — a function can be continuous without being differentiable, but if a function is differentiable, it is also continuous.

For the given function \(f(x) = 2x-2-\cos x\), it is differentiable everywhere because it is composed of polynomial and trigonometric parts, both of which are known to be differentiable. The derivative of the function, \(f'(x) = 2 + \sin x\), suggests how the function's rate of change behaves, which is crucial for applying Rolle's Theorem to prove the uniqueness of a solution as demonstrated in the exercise.

Continuity of Functions

Continuity is a fundamental property of functions in calculus. A function is said to be continuous at a point if the limit of the function as it approaches the point is equal to the function's value at that point. For a function to be continuous on a closed interval, it must be continuous at every point within that interval.

The concept is essential in applying the Intermediate Value Theorem, which asserts that if a function is continuous on a closed interval \[a, b\] and takes on different values at two points, then it must take on every value between those two points at least once. In the context of our exercise, the continuity of \(f(x) = 2x-2-\cos x\) over the closed interval \[0, 1\] means the function crosses the x-axis somewhere between \(x = 0\) and \(x = 1\), ensuring the existence of at least one real solution to the equation.