Problem 58 - University Calculus: Early Transcendentals

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 \(\quad f(x)=1 / x\) on an interval of positive numbers \([a, b]\) is \(c=\sqrt{a b}\).

Step-by-Step Solution

Answer

Thus, \(c = \sqrt{ab}\) is the value satisfying the Mean Value Theorem for \(f(x) = \frac{1}{x}\).

Step 1: Understanding the Mean Value Theorem

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

Step 2: Define the function and find its derivative

Given the function f(x) = \frac{1}{x}, we first find its derivative \[ f'(x) = -\frac{1}{x^2} \].

Step 3: Apply the Mean Value Theorem formula

Using the definition from the Mean Value Theorem:\[ f'(c) = \frac{f(b) - f(a)}{b-a} \]We have:\[ -\frac{1}{c^2} = \frac{\frac{1}{b} - \frac{1}{a}}{b-a} \]

Step 4: Simplify the expression on the right

The expression \(\frac{1}{b} - \frac{1}{a}\) can be simplified as:\[ \frac{a - b}{ab} \]Thus, the Mean Value Theorem equation becomes:\[ -\frac{1}{c^2} = \frac{a - b}{ab(b-a)} \] which simplifies to \[ -\frac{1}{c^2} = \frac{1}{ab} \]

Step 5: Solve for \(c\)

Rearranging the equation \[ -\frac{1}{c^2} = \frac{1}{ab} \] gives us:\[ c^2 = ab \]Taking the square root of both sides, we find \[ c = \sqrt{ab} \].

Key Concepts

Geometric Mean

When we're dealing with numbers, the geometric mean offers a way to find a middle value that is more appropriate for certain types of data. It's especially handy when we deal with proportions or percentages. To calculate the geometric mean of two positive numbers, say \( a \) and \( b \), we take the square root of their product: \( \sqrt{a \cdot b} \). This value falls between \( a \) and \( b \) if both numbers are the same or closely related.

• The geometric mean is particularly useful in situations where data are products rather than sums.
• It helps to find a central tendency in a multiplicative context, like rates of growth.

In the context of our exercise, the geometric mean provides the value \( c \) in the Mean Value Theorem, demonstrating a unique intersection between mathematical concepts.

Derivative

The derivative of a function is a concept from calculus that represents the rate of change or slope of the function. If you're ever in need of understanding how something changes instantaneously, derivatives come into play. Mathematically speaking, it involves finding the limit of the average rate of change as the interval approaches zero. For our function \( f(x) = \frac{1}{x} \), the derivative is \( f'(x) = -\frac{1}{x^2} \).

• Derivatives help in understanding motion, growth rates, and many real-world phenomena.
• They provide critical information for optimization problems, where you want to find maxima or minima.

In our exercise, the derivative is critical to applying the Mean Value Theorem and finding the precise point \( c \) where the theorem holds.

Continuous Function

A continuous function is like a smooth path without any jumps or breaks. In mathematical terms, a function \( f(x) \) is continuous on an interval \([a, b]\) if it is defined at every point in that interval and does not have any discontinuities. Think about a continuous line that you can draw without ever lifting your pencil.

• Continuity ensures that function values change gradually, not abruptly.
• Functions that are continuous over an interval can be used reliably for interpolation and estimating unknown values.

For the Mean Value Theorem to be applicable, the function must be continuous on the closed interval \([a, b]\). This ensures the function behaves predictably over the entire range.

Differentiable Function

A differentiable function is one that has a derivative at each point in its domain. This means there's a well-defined tangent line at each point, allowing us to discuss the rate of change at any point. For a function to be differentiable on an interval, it must be smooth without any sharp turns or corners.

• Differentiability implies continuity, but the reverse is not true.
• It's a stronger condition than continuity, requiring the function to be "smooth" in its behavior.

In our exercise, differentiability on the open interval \((a, b)\) is required for the Mean Value Theorem to apply, ensuring a valid \( c \) exists where the derivative equals the average rate of change over the interval.