Problem 15 - Calculus with Concepts in Calculus

Question

Suppose \(\left|f^{\prime}(x)\right| \leq M\) for \(a \leq x \leq b\). Using the Mean Value Theorem, prove that \(|f(b)-f(a)| \leq M(b-a)\), so that \(f(a)-M(b-a) \leq f(b) \leq f(a)+M(b-a)\).

Step-by-Step Solution

Answer

Using MVT, \( |f(b) - f(a)| \leq M(b-a) \) because \( |f'(c)| \leq M \).

Step 1: Recall the Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function \( f(x) \) is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists a point \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Step 2: Apply the Mean Value Theorem

By the MVT, there exists some point \( c \in (a, b) \) where \( f'(c) = \frac{f(b) - f(a)}{b - a} \). This means the derivative \( f'(x) \) at some point \( c \) approaches the average rate of change of the function over the interval \([a, b]\).

Step 3: Relate the Derivative to the Given Condition

We are given that \( |f'(x)| \leq M \) for all \( x \in [a, b] \). In particular, this means \( |f'(c)| \leq M \).

Step 4: Establish the Inequality

Since \( f'(c) = \frac{f(b) - f(a)}{b - a} \), applying the absolute value we get \( \left| \frac{f(b) - f(a)}{b - a} \right| \leq M \).

Step 5: Simplify the Inequality

By multiplying both sides of \( \left| \frac{f(b) - f(a)}{b - a} \right| \leq M \) by \( |b-a| \), we have \( |f(b) - f(a)| \leq M |b-a| \).

Step 6: Conclude the Result

Since \( b \geq a \), we have \( b-a = |b-a| \). Therefore, \( |f(b) - f(a)| \leq M(b-a) \). This results in the inequality \( f(a) - M(b-a) \leq f(b) \leq f(a) + M(b-a) \).

Key Concepts

calculus

Calculus is a branch of mathematics that deals with the concepts of change and motion. It provides tools to analyze functions and their behavior. Two primary operations in calculus are differentiation and integration.

Differentiation involves finding the derivative of a function, which represents the rate of change of the function with respect to a variable. In simpler terms, it tells us how a function is changing at any given point. This concept is crucial when we discuss the Mean Value Theorem, as it forms the basis for understanding how a function behaves between two points.

Integration, on the other hand, is the process of finding the integral of a function, which can be thought of as the opposite of differentiation. It helps in finding areas under curves, among other applications.

• The Mean Value Theorem (MVT), a fundamental theorem in calculus, connects differentiation with the average rate of change of a function.
• MVT asserts that, given a function continuous on \([a, b]\) and differentiable on \( (a, b) \), there exists at least one point where the instantaneous rate of change (derivative) matches the average rate of change over \([a, b]\).

Understanding calculus is critical for solving problems related to the behavior of functions, like the inequality demonstrated in our exercise.

inequality proof

An inequality proof is a mathematical demonstration showing that one quantity is less than, greater than, or similar to another. In this exercise, we aimed to prove an inequality using the Mean Value Theorem.

The problem stated aims to show \(|f(b) - f(a)| \leq M(b-a)\) given that \(|f'(x)| \leq M\). Validating this involves several steps that ensure our conclusions are based on solid mathematical grounding.

• First, we apply the Mean Value Theorem, which implies the existence of a point \(c \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
• Recognize that since \(|f'(x)| \leq M\) holds for all \(x\), it must also hold for \(c\).
• Hence, use the inequality \(| f'(c) | \leq M\) to establish \(|\frac{f(b) - f(a)}{b - a}| \leq M\).
• Finally, multiply through by \(b-a\) to yield the desired \(|f(b)-f(a)| \leq M(b-a)\).

This chain of logical steps is crucial in demonstrating the relationship between the change in the function's value and its derivative, leading to the conclusive inequality.

differentiability

Differentiability is a property of a function that indicates it has a well-defined derivative at each point in a certain interval. For a function to be differentiable at a point, it must be smooth and have no corners, cusps, or discontinuities at that point.

The significance of differentiability in the context of the Mean Value Theorem is paramount. The theorem requires that the function is differentiable on the open interval \( (a, b) \). This is because the conclusion of the theorem hinges on the existence of a derivative that reflects how the function transitions between two endpoints, \(a\) and \(b\).

• If a function is not differentiable at any point within the interval, the Mean Value Theorem doesn't apply, as it relies on the consistency of the derivative to find a specific point \(c\).
• Differentiability ensures the function doesn't have abrupt changes, allowing us to use derivatives meaningfully to figure out the rate of change.

The property of differentiability is what enables mathematicians to make predictions about the behavior of functions over intervals, such as those seen in this exercise.

continuity

Continuity is a foundational concept that describes a function that is unbroken and without interruptions over a domain. A function is continuous if you can draw its graph without lifting your pencil off the paper. This smoothness is crucial for many theorems in calculus, including the Mean Value Theorem.

The Mean Value Theorem requires the function to be continuous on the closed interval \([a, b]\). Here’s why continuity matters:

• Continuity ensures that there are no jumps or breaks in the function's graph. This is essential as any discontinuity means there could be an undefined or infinite derivative, violating the conditions of the Mean Value Theorem.
• The presence of continuity guarantees that the function possesses a value at every point within \([a, b]\), ensuring the transition across the interval is smooth.

Without continuity, there can be no assurance of the existence of the specific point \(c\) dictated by the Mean Value Theorem. Thus, in this exercise, continuity is just as vital as differentiability to establish the given inequalities using the theorem.