Problem 50 - Single Variable Calculus

Question

In Exercises \(47-52\), determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. If \(\left|f^{\prime}(x)\right| \leq 1\) for all \(x\), then $$ \left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leq\left|x_{1}-x_{2}\right| $$ for all numbers \(x_{1}\) and \(x_{2}\).

Step-by-Step Solution

Answer

The given statement is true. If the absolute value of the derivative of a function, \(\left|f'(x)\right|\), is less than or equal to 1 for all \(x\), then the absolute difference of function values at any two points, \(\left|f(x_1) - f(x_2)\right|\), is less than or equal to the absolute difference of these two points, \(\left|x_1 - x_2\right|\). The Mean Value Theorem is used to prove this statement.

Step 1: Understand the given statement

The statement we need to verify is: "If \(\left|f^{\prime}(x)\right| \leq 1\) for all \(x\), then \[\leq\left|x_{1}-x_{2}\right|\] for all numbers \(x_{1}\) and \(x_{2}\)." What the statement asserts is that if the absolute value of the derivative of a function is always less than or equal to 1, then the absolute difference of function values at any two points is less than or equal to the absolute difference of these two points.

Step 2: Use Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function is differentiable on the open interval \((a,b)\) and continuous on the closed interval \([a,b]\), then there exists a point \(c\) in \((a,b)\) such that \[f'(c) = \frac{f(b) - f(a)}{b - a}\] Now, let's apply MVT to the given problem. Let \(f\) be differentiable on \((x_1,x_2)\) and continuous on \([x_1,x_2]\). According to MVT, there exists a point \(c\) in \((x_1,x_2)\) such that \[f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}\] We know that \(\left|f^{\prime}(x)\right| \leq 1\) for all \(x\), which means that \(-1 \le f'(c) \le 1\). Multiplying by \((x_2 - x_1)\), we get: \[-(x_2 - x_1) \le f(x_2) - f(x_1) \le (x_2 - x_1)\] Since the given statement is concerned with absolute values, we need to rewrite this inequality in terms of absolute values: \[\left|f(x_2) - f(x_1)\right| \le \left|(x_2 - x_1)\right|\] Therefore, the given statement is true, and the explanation is that if the absolute value of the derivative of a function is always less than or equal to 1, the Mean Value Theorem proves that the absolute difference of function values at any two points will be less than or equal to the absolute difference of these two points.

Key Concepts

Derivative of a Function

Understanding the derivative of a function is a foundational aspect of differential calculus. The derivative can be seen as the rate at which a function's output value changes in response to changes in its input value. In other words, it measures how a function's value will adjust if we were to slightly adjust the input, which we can think of as looking at the slope of the function at a particular point.

Imagine driving a car and the speedometer is giving you the speed at a specific instant—that's like the derivative: it tells you how fast you're going right now, not in the past, or the future, but at this instant. Mathematically, if we denote the function as \( f(x) \), its derivative is written as \( f'(x) \) or \( \frac{df}{dx} \), where \( x \) is the input variable and \( f'(x) \) represents how 'quickly' or 'slowly' the function's value is changing at each point.

Let's say you have a function that models the elevation of a hill based on the distance you've walked. The derivative at a particular point on the hill will tell you how steep the hill is right there. A steep hill will have a high derivative value (either positive or negative depending on the direction), while a flat part will have a derivative value close to zero, indicating very little change.

Absolute Value

The absolute value of a real number is a non-negative value without regard to its sign. Simply put, it measures the 'distance' a number is from zero on the number line, so it's always positive. For any real number \( x \), its absolute value is denoted as \( |x| \), and it equals \( x \) if \( x \) is greater than or equal to zero, and \( -x \) if \( x \) is less than zero.

Imagine you and a friend are on a number line: you're standing on \( 3 \) and your friend is standing on \( -3 \). Despite being in opposite directions, you're both exactly three steps away from the starting point at zero. That's how absolute value works: it's about the distance, not the direction.

In the context of calculus, absolute values can sometimes complicate things because they make us consider both the positive and negative scenarios. For instance, when we take the absolute value of a function's derivative, we're interested in how steep the function is, regardless of the direction of the slope—whether it's going uphill or downhill.

Differential Calculus

Differential calculus is a major branch of calculus that deals with the concept of differentiation, which allows us to analyze the rates at which things change. This process can help us understand a wide range of phenomena, from the simple motion of objects to the behavior of complex functions in engineering and science.

When studying differential calculus, we often talk about finding the derivative of a function – a process known as differentiation. It is through differentiation that we can dissect the function's behavior, find maximum and minimum values, determine points of inflection, and see where a function increases or decreases.

Consider a balloon that's being filled with air. Differential calculus can tell us not just how much the balloon has inflated over time but also the rate of its inflation at any moment. If we plot the balloon's volume on a graph over time, the derivative at each point will tell us how quickly that volume is changing then and there. Understanding this branch of calculus is crucial when any rate of change is involved, and it's the basis for a multitude of applications in science, economics, engineering, and beyond.