Problem 55 - Thomas Calculus in SI Units

Question

The geometric mean of two positive numbers \(a\) and \(b\) is the number \(\sqrt{a b}\). Show that the value of \(c\) in the conclusion of the Mean Value Theorem for \(f(x)=1 / x\) on an interval of positive numbers \([a, b]\) is \(c=\sqrt{a b}\)

Step-by-Step Solution

Answer

The value of \( c \) is \( \sqrt{ab} \).

Step 1: Statement of the Mean Value Theorem

The Mean Value Theorem (MVT) states that for a continuous function \( f \) on a closed interval \( [a, b] \) and differentiable on the open interval \( (a, b) \), there exists at least one \( c \) in \( (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b-a} \).

Step 2: Define the Function and Calculate Values

Given \( f(x) = \frac{1}{x} \), we evaluate \( f(a) = \frac{1}{a} \) and \( f(b) = \frac{1}{b} \). So the difference quotient is \( \frac{f(b) - f(a)}{b-a} = \frac{\frac{1}{b} - \frac{1}{a}}{b-a} \).

Step 3: Simplify the Difference Quotient

Simplifying the difference quotient gives \( \frac{1}{b} - \frac{1}{a} = \frac{a-b}{ab} \). Thus, the difference quotient becomes \( \frac{\frac{a-b}{ab}}{b-a} = -\frac{1}{ab} \).

Step 4: Compute the Derivative of the Function

The derivative of \( f(x) = \frac{1}{x} \) is \( f'(x) = -\frac{1}{x^2} \).

Step 5: Set Up the MVT Equation

According to MVT, \( f'(c) = \frac{f(b) - f(a)}{b-a} \) implies \( -\frac{1}{c^2} = -\frac{1}{ab} \).

Step 6: Solve for \(c\)

Equating \( -\frac{1}{c^2} = -\frac{1}{ab} \) gives \( c^2 = ab \). Solving for \( c \), we have \( c = \sqrt{ab} \).

Key Concepts

Geometric Mean

The geometric mean is a special type of average that is quite different from the traditional arithmetic mean. It is the number that represents the central tendency or typical value of a set of numbers. For two positive numbers, the geometric mean is found by taking the square root of their product.

For example, if you have two numbers, \( a \) and \( b \), the geometric mean is calculated as:

• \( \text{Geometric Mean} = \sqrt{ab} \)

This concept is not just crucial in mathematics, but also used in various fields such as finance and statistics. It provides a way to compare different quantities with different ranges by normalizing them, helping to provide a balance to the data.
In the context of the Mean Value Theorem, finding the geometric mean helps identify the specific point \( c \) at which the average rate of change equals the instantaneous rate of change of the given function.

Derivative

A derivative is a fundamental concept in calculus that describes how a function changes as its input changes. It measures the rate at which a quantity changes. The derivative of a function \( f(x) \) at a point \( x \) is the slope of the tangent line to the function at that point.

The derivative of the function \( f(x) = \frac{1}{x} \) is calculated as:

• \( f'(x) = -\frac{1}{x^2} \)

This tells us that the slope of the tangent at any point on the curve of \( f(x) = \frac{1}{x} \) is negative and inversely proportional to the square of \( x \).

Understanding derivatives is essential because they enable us to find rates of change and identify maximum and minimum values of functions. In the context of the Mean Value Theorem, we need the derivative of the function to find the value \( c \) where the tangent is parallel to the secant line connecting two points of the function.

Difference Quotient

The difference quotient is key to understanding the concept of derivatives. It is the formula used to find the average rate of change of the function over a specific interval before taking the limit to derive the instantaneous rate of change or derivative.

For a function \( f(x) \), the difference quotient between points \( a \) and \( b \) is given by:

• \( \frac{f(b) - f(a)}{b-a} \)

In our specific example with the function \( f(x) = \frac{1}{x} \), we calculated:

• \( \frac{\frac{1}{b} - \frac{1}{a}}{b-a} = \frac{a-b}{ab(b-a)} = -\frac{1}{ab} \)

This result represents the average rate of change over the interval \([a, b]\). The Mean Value Theorem then ensures that there exists a point \( c \) within the interval where the instantaneous rate of change (derivative) equals this average rate of change represented by the difference quotient.