Problem 12 - Essential Calculus Early Transcendentals

Question

Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers \(c\) that satisfy the conclusion of the Mean Value Theorem. \(f(x)=1 / x, \quad[1,3]\)

Step-by-Step Solution

Answer

The function satisfies the hypotheses, and the number \(c\) is \(\sqrt{3}\).

Step 1: Check Continuity

The Mean Value Theorem requires that the function be continuous on the closed interval \( [a, b] \). For the function \( f(x) = \frac{1}{x} \), it is continuous everywhere except where its denominator is zero. In the interval \([1, 3]\), \( f(x) \) is continuous because the denominator is non-zero throughout (1 and 3 are both non-zero). Therefore, \(f(x) = \frac{1}{x}\) is continuous on \([1, 3]\).

Step 2: Check Differentiability

The Mean Value Theorem also requires the function to be differentiable on the open interval \( (1, 3) \). The derivative of \( f(x) = \frac{1}{x} \) is \( f'(x) = -\frac{1}{x^2} \), which is defined and continuous for all \( x eq 0 \). Thus, \( f(x) = \frac{1}{x} \) is differentiable on \( (1, 3) \).

Step 3: Apply Mean Value Theorem

Since \( f(x) = \frac{1}{x} \) is continuous on \([1, 3]\) and differentiable on \( (1, 3) \), the Mean Value Theorem applies. The theorem states that there exists some \(c \) in \( (1, 3) \) such that \( f'(c) = \frac{f(3) - f(1)}{3 - 1}.\)

Step 4: Calculate \\(f'(c)\\)

Substitute \( f(x) = \frac{1}{x} \) and its derivative \( f'(x) = -\frac{1}{x^2} \) into the Mean Value Theorem equation:\(-\frac{1}{c^2} = \frac{\frac{1}{3} - 1}{2} = \frac{-\frac{2}{3}}{2} = -\frac{1}{3}.\)Now, solve for \( c \).

Step 5: Solve for \\( c \\\)

Set \( -\frac{1}{c^2} = -\frac{1}{3} \) and solve for \( c \):\( c^2 = 3 \), which gives \( c = \sqrt{3} \).Since \(c = \sqrt{3} \) lies within the interval \( (1, 3) \), it satisfies the Mean Value Theorem.

Key Concepts

Continuity

Continuity is a fundamental property of functions that is crucial in calculus. A function is continuous on a closed interval if it has no breaks, jumps, or holes over that interval.
For the Mean Value Theorem to apply, a function must be continuous over a closed interval \( [a, b] \). This means the function should smoothly transition from one point to another within the interval without any interruptions.
For example, in the function \( f(x) = \frac{1}{x} \), the need for continuity means we must ensure that the denominator \( x \) never becomes zero because division by zero is undefined. In the interval \([1, 3]\), \( f(x) = \frac{1}{x} \) is continuous since the denominator remains positive, providing consistent and defined values throughout.

• Continuous functions are predictable and smooth.
• Checking for continuity ensures no unexpected drops or gaps in the graph.

Differentiability

Differentiability is another critical property that a function must satisfy for the Mean Value Theorem to hold. A function is differentiable at a point if it has a defined tangent at that point. Essentially, the function must be smooth and not form any sharp corners or cusps.
For a function to be differentiable on an open interval \( (a, b) \), it must have a derivative that is defined at every point within this interval. The derivative represents the rate at which the function's value is changing, akin to the slope of the function at any given point.
In the case of \( f(x) = \frac{1}{x} \), its derivative \( f'(x) = -\frac{1}{x^2} \) is defined and continuous when \( x \) is not zero. Therefore, over the open interval \( (1, 3) \), \( f(x) \) remains differentiable as \( x \) does not hit a zero value point.

• Differentiable functions have no sharp edges.
• Being differentiable indicates a smoothly changing gradient.

Derivatives

Derivatives are at the heart of calculus and are especially important in understanding and applying the Mean Value Theorem. A derivative represents the slope of the tangent line to the function at a given point, indicating how fast the function’s value is changing relative to changes in the input.
In mathematical symbols, the derivative of a function \( f(x) \) is often denoted as \( f'(x) \) or \( \frac{df}{dx} \). Derivatives provide insight into the behavior of functions, showcasing where they increase, decrease, or maintain constant values.
For \( f(x) = \frac{1}{x} \), the derivative is \( f'(x) = -\frac{1}{x^2} \). This derivative is negative over its domain, implying that the function is decreasing wherever it is defined.

• Derivatives help us identify slopes of tangent lines.
• They offer a way to predict function behavior over intervals.
• A negative derivative shows a function decreasing.

Interval Analysis

Interval analysis involves examining specific segments or intervals of the domain of a function to apply mathematical theorems or properties effectively. For the Mean Value Theorem, it means analyzing the intervals where the function satisfies the required conditions.
The Mean Value Theorem requires a function to be continuous on a closed interval \( [a, b] \) and differentiable on the open interval \( (a, b) \). This careful analysis ensures the theorem can be applied correctly, allowing us to find specific points within the interval that satisfy certain conditions.
In our example, with the function \( f(x) = \frac{1}{x} \), interval analysis shows that the function is continuous and differentiable over \([1, 3]\). This verifies that the Mean Value Theorem can be applied here, leading to the conclusion that there exists a point \( c \) such that \( f'(c) = \frac{f(3) - f(1)}{3-1} \).

• Interval analysis is important for applying calculus theorems.
• It helps in identifying applicable intervals for continuity and differentiability.
• Proper interval analysis ensures mathematical proofs and applications are accurate.