Problem 60 - Calculus with Concepts in Calculus

Question

a. Let \(f(x)=x-\sin x\). Show that \(f\) is increasing. b. Use part (a) to show that \(\sin x>x\) on \((-\infty, 0)\) and \(\sin x

Step-by-Step Solution

Answer

\(f(x) = x - \sin x\) is increasing because \(1 - \cos x \geq 0\), hence for \(x<0\), \(\sin x > x\) and for \(x>0\), \(\sin x < x\).

Step 1: Understanding the Problem

We need to show that the function \(f(x) = x - \sin x\) is increasing. A function is increasing if its derivative is positive everywhere on its domain.

Step 2: Differentiate the Function

Find the derivative of \(f(x)\). The derivative of \(x\) is 1, and the derivative of \(-\sin x\) is \(-\cos x\). Thus, \(f'(x) = 1 - \cos x\).

Step 3: Analyze the Sign of the Derivative

Determine where \(f'(x) = 1 - \cos x > 0\). Since \(-1 \leq \cos x \leq 1\), this implies \(1 - \cos x \geq 0\), with equality only at points where \(\cos x = 1\), which occurs when \(x = 2k\pi\), where \(k\) is an integer.

Step 4: Conclude that \(f(x)\) is Increasing

Since \(f'(x) = 1 - \cos x\) is non-negative and is greater than zero except at isolated points \(x = 2k\pi\), \(f(x)\) is increasing on the entire real line.

Step 5: Use the Result to Compare \(\sin x\) and \(x\)

Using that \(f(x) = x - \sin x\) is increasing, consider the signs at a point. - For \(x = 0\), \(f(0) = 0 - \sin 0 = 0\).- For \(x < 0\), since \(f(x)\) is increasing and \(f(0) = 0\), \(f(x) < 0\) implying \(x - \sin x < 0\) or \(\sin x > x\) on \((-ifty, 0)\).- For \(x > 0\), similarly, \(f(x) > 0\) implying \(\sin x < x\) on \((0, \infty)\).

Key Concepts

Derivatives

In calculus, a derivative represents the rate at which a function is changing at any given point. It's like looking at the slope of a curve at a particular point. For the function \(f(x) = x - \sin x\), the derivative \(f'(x)\) tells us how \(f(x)\) changes as \(x\) changes.

The process of finding the derivative is called differentiation. For \(f(x) = x - \sin x\), we compute the derivative by finding the derivative of each part separately. The derivative of \(x\) with respect to \(x\) is 1, since it's a linear function. Next, the derivative of \(-\sin x\) is \(-\cos x\), as the derivative of \(\sin x\) is \(\cos x\).

Thus, the derivative of the function is \(f'(x) = 1 - \cos x\). Understanding derivatives is crucial in analyzing how functions behave over their domains.

Trigonometric Functions

Trigonometric functions like \(\sin x\) and \(\cos x\) are foundational in mathematics, especially in the study of waves, circles, and oscillations. These functions are periodic, with cycles repeating at regular intervals.

For \(\sin x\) and \(\cos x\), their values range between -1 and 1. Notably, \(\cos x\) reaches 1 when \(x\) equals any multiple of \(2\pi\), such as \(0, 2\pi, 4\pi,\) and so on. This periodic behavior is central to solving problems involving trigonometric functions.

In our original problem, understanding these ranges and properties of \(\cos x\) helped determine when \(1 - \cos x > 0\), which indicates that the function \(f(x) = x - \sin x\) is increasing.

Differentiation

Differentiation is the mathematical process of finding a derivative. It's key for analyzing how functions change and behave. By differentiating a function, you can understand something about its rate of change and how it varies across its domain.

In the context of the function \(f(x) = x - \sin x\), we applied differentiation to find \(f'(x) = 1 - \cos x\). The derivative tells us more about the function's behavior. In particular, it lets us infer where the function is increasing or decreasing based on the sign of the derivative.

This technique is particularly valuable in calculus because it transforms complex problems into simpler ones. By understanding how to perform differentiation and interpret derivatives, solving problems becomes more manageable.

Monotonicity

Monotonicity refers to the property of a function being entirely non-decreasing or non-increasing. In simpler terms, a function is monotonic if it moves in the same direction across its domain.

For \(f(x) = x - \sin x\), we examined the derivative \(f'(x) = 1 - \cos x\) to determine the function's monotonicity. Since \(\cos x\) ranges between \(-1\) and \(1\), it follows that \(1 - \cos x\) is always positive or zero except at specific points, like \(0, 2\pi, 4\pi\), where it equals zero.

This result shows that \(f(x)\) is a monotonically increasing function. On the entire real line, \(f(x)\) consistently rises, helping us conclude that on intervals like \((-\infty, 0)\) and \((0, \infty)\), \(\sin x\) behaves predictably in relation to \(x\). Such analyses are practical in understanding the broader behavior of mathematical models.