Problem 52 - Thomas Calculus

Question

The arithmetic mean of \(a\) and \(b\) The arithmetic mean of two numbers \(a\) and \(b\) is the number \((a+b) / 2 .\) Show that the value of \(c\) in the conclusion of the Mean Value Theorem for \(f(x)=x^{2}\) on any interval \([a, b]\) is \(c=(a+b) / 2\).

Step-by-Step Solution

Answer

The value of \( c \) is \( \frac{a+b}{2} \).

Step 1: Understanding the Mean Value Theorem

The Mean Value Theorem states that for a continuous function \( f \) differentiable on the open interval \((a, b)\), there exists a point \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). Here, \( f(x) = x^2 \). Our goal is to find \( c \) in terms of \( a \) and \( b \).

Step 2: Calculating the Derivative

First, find the derivative of \( f(x) = x^2 \). The derivative \( f'(x) = 2x \). We'll use this to find the value of \( c \).

Step 3: Apply the Mean Value Theorem Formula

According to the Mean Value Theorem, we need to set \( f'(c) = \frac{f(b) - f(a)}{b - a} \). So we have \( 2c = \frac{b^2 - a^2}{b-a} \).

Step 4: Simplifying the Right Side

Simplify \( \frac{b^2 - a^2}{b-a} \) using the difference of squares formula: \( b^2 - a^2 = (b-a)(b+a) \). This gives us \( \frac{b^2 - a^2}{b-a} = b + a \).

Step 5: Solve for c

Equate \( 2c = b + a \) from the last step. Solve for \( c \): \[ c = \frac{b + a}{2} \]. Therefore, \( c = \frac{a + b}{2} \).

Key Concepts

Arithmetic Mean

The arithmetic mean, often simply called the average, is a fundamental concept in math. It provides a way to find the central value of a set of numbers. For two numbers, say \( a \) and \( b \), the arithmetic mean is calculated using the formula \( \frac{a+b}{2} \).
This represents the midpoint between \( a \) and \( b \), offering a balance point on the number line. When you think about averages, it's a practical tool in everyday life:

• It helps determine the central tendency of data.
• It's used in various fields, from finance to engineering.
• In this exercise, the arithmetic mean plays a crucial role, revealing the special point \( c \) that satisfies the Mean Value Theorem for our function.

Understanding the arithmetic mean is crucial, especially when solving problems involving specific average values, like the one in the Mean Value Theorem.

Derivative

A derivative is a powerful mathematical concept that describes the rate at which a function is changing at any given point. Essentially, it measures how a function's output value shifts as the input value changes.
For the function \( f(x) = x^2 \) used in this problem:

• The derivative, notated as \( f'(x) \), is calculated as \( 2x \).
• This tells us that the slope of the tangent line to the curve \( x^2 \) at any point \( x \) is \( 2x \).
• In the Mean Value Theorem context, the derivative helps pinpoint the exact 'average' slope on the interval \([a, b]\).

By understanding derivatives, you unlock a deeper insight into how functions behave, anticipate changes, and solve complex calculus problems. They are a cornerstone for comprehending not only this exercise but a broad range of mathematical and real-world applications.

Continuous Functions

A continuous function is a function that smoothly progresses without any interruptions or jumps. In algebra and calculus, these are crucial because they allow for clearer analysis and prediction of behavior.
In terms of the Mean Value Theorem, a continuous function on a closed interval \([a, b]\) ensures:

• The function has no breaks or gaps in its graph from \( a \) to \( b \).
• There exists at least one point \( c \) where the derivative equals the average slope between \( a \) and \( b \).
• A key requirement for applying the theorem is the function must be continuous on the interval and differentiable on \((a, b)\).

For \( f(x) = x^2 \), it is continuous (since it forms a smooth parabola) and differentiable over any real interval. This property is why we can confidently apply the Mean Value Theorem to determine the point \( c \), showing how continuous functions underpin foundational theorems in calculus.

Difference of Squares

The difference of squares is a useful algebraic identity that helps break down expressions for simplification or factorization. It states that for any two numbers \( a \) and \( b \), the difference of their squares is given by:

• \( a^2 - b^2 = (a-b)(a+b) \)

In this exercise, this identity was key to simplifying the expression \( \frac{b^2 - a^2}{b-a} \). By recognizing it as a difference of squares, we can factor it:

• Substitute \( b^2 - a^2 \) with \((b-a)(b+a)\).
• Cancel out the \( (b-a) \) terms to get \( b+a \).
• This simplification made it straightforward to solve for \( c = \frac{b+a}{2} \).

Mastering the difference of squares identity offers a powerful tool in algebra for making complex expressions simpler and solving them with ease. It's not just pivotal in textbook problems but also in many practical scenarios where equation transformation is needed.