Problem 55 - Thomas Calculus

Question

The geometric mean of \(a\) and \(b\) . 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: Understand the Mean Value Theorem

The Mean Value Theorem states that for a function that is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists a number \(c\) in \((a, b)\) such that \[f'(c) = \frac{f(b) - f(a)}{b - a}.\] We'll find \(c\) for the function \(f(x) = \frac{1}{x}\).

Step 2: Calculate the derivative of the function

For \(f(x) = \frac{1}{x}\), the derivative \(f'(x)\) is found using the power rule or reciprocal rule, obtaining:\[f'(x) = -\frac{1}{x^2}.\]

Step 3: Apply the Mean Value Theorem

Set the derivative equal to the difference quotient as per the Mean Value Theorem:\[-\frac{1}{c^2} = \frac{\frac{1}{b} - \frac{1}{a}}{b - a}.\]

Step 4: Simplify the right side of the equation

Simplify \(\frac{\frac{1}{b} - \frac{1}{a}}{b - a}\):\[\frac{1}{b} - \frac{1}{a} = \frac{a - b}{ab},\]so the expression becomes:\[\frac{\frac{1}{b} - \frac{1}{a}}{b - a} = \frac{a-b}{ab(b-a)} = -\frac{1}{ab}.\]

Step 5: Solve for \(c\)

Equating both sides results in:\[-\frac{1}{c^2} = -\frac{1}{ab}.\]Removing the negative sign, we have:\[\frac{1}{c^2} = \frac{1}{ab}.\]Taking reciprocals, we get:\[c^2 = ab.\]Finally, taking the square root of both sides, \[c = \sqrt{ab}.\]

Key Concepts

Geometric Mean

The geometric mean is a central concept in mathematics, particularly when dealing with growth rates or ratios of quantities. It is represented as \( \sqrt{ab} \,\) for two positive numbers \(a\) and \(b\), and provides a way to find a central tendency that is more appropriate than the arithmetic mean when dealing with products or exponential growth.

To understand it better, think of the geometric mean as the side length of a square whose area is the same as the rectangle's area with sides \(a\) and \(b\). This makes it very insightful in geometry and certain financial calculations.

• The geometric mean is especially useful in proportional growth situations, such as population growth or financial investments.
• It gives a measure that balances high and low values, more representative than a simple average.
• In our exercise, finding \(c = \sqrt{ab}\) using the Mean Value Theorem illustrates how geometric principles apply to calculus.

Differentiable Functions

Differentiability is fundamental in calculus, reflecting whether a function has a derivative at each point. For a function to be differentiable, it must be smooth without any breaks, corners, or cusps.

In the Mean Value Theorem, we apply it on a function \(f(x) = \frac{1}{x}\), which is both continuous and differentiable on its domain—except where \(x = 0\). Because the interval \([a, b]\) contains only positive values, differentiability is assured throughout this range.

• A differentiable function allows for the calculation of an instantaneous rate of change, which is the derivative.
• Mean Value Theorem leverages differentiability to establish a specific point \(c\) where the function's instantaneous rate of change matches the function's average rate of change over \([a, b]\).
• Differentiability implies continuity but not vice versa, so always verify function smoothness when using calculus theorems.

Derivative Calculation

The calculation of a derivative is a cornerstone of calculus, revealing how a function changes at any given point. In our exercise, we handle the derivative of \(f(x) = \frac{1}{x}\), finding it through the rules of differentiation.

Using the power rule or reciprocal rule, we derive: \( f'(x) = -\frac{1}{x^2}\). This step highlights the procedure of finding a derivative, which provides insights into the curve's slope for any \(x\).

• The power rule states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\), and in reciprocal fashion, \(x^{-1}\) transforms the formula accordingly.
• Understanding the derivative's form helps in predicting the behavior of \(f(x)\) at specific intervals, clarifying the changes.
• In tandem with Mean Value Theorem, these calculations provide the foundations for solving real-world rate-of-change problems and connecting them with geometry.