Problem 67 - Thomas Calculus in SI Units

Question

Assume that \(f\) is differentiable on \(a \leq x \leq b\) and that \(f(b)

Step-by-Step Solution

Answer

By MVT, since \( f(b) < f(a) \), there exists a point \( c \) in \((a, b)\) such that \( f'(c) < 0 \).

Step 1: Understand the Problem

The function \( f \) is differentiable on the interval \( [a, b] \), and it is given that \( f(b) < f(a) \). Our task is to show there exists a point \( c \) in the interval \( (a, b) \) where the derivative of \( f \), \( f'(c) \), is negative.

Step 2: Apply Rolle's Theorem Preconditions

Recall the Mean Value Theorem (MVT), which requires a function to be continuous on \([a, b]\) and differentiable on \((a, b)\). Here, \( f \) is differentiable on \( [a, b] \), so these conditions are satisfied.

Step 3: Apply Mean Value Theorem (MVT)

The Mean Value Theorem states that if \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Step 4: Analyze the Slope of the Secant Line

The slope of the secant line joining points \( (a, f(a)) \) and \( (b, f(b)) \) is given by \( \frac{f(b) - f(a)}{b - a} \). Since \( f(b) < f(a) \), it follows that this slope is negative: \( \frac{f(b) - f(a)}{b - a} < 0 \).

Step 5: Conclude from MVT

Since the slope \( \frac{f(b) - f(a)}{b - a} \) is negative, the Mean Value Theorem guarantees there exists a point \( c \) in \( (a, b) \) such that \( f'(c) \) equals this negative slope. Hence, \( f'(c) < 0 \).

Key Concepts

Differentiability

Differentiability is a crucial concept in calculus. It tells us if a function has a derivative at each point in its domain. For a function to be differentiable at a point, it must be smooth around that point without any sharp edges or discontinuities.
A differentiable function is also continuous, meaning that you can draw the curve without lifting your pencil off the paper. This implies that the rate of change (derivative) is well-defined.
In our problem, the function \( f \) is differentiable over the interval \([a, b]\). This means that everywhere between \(a\) and \(b\), the function is smooth, and we can compute a derivative, \( f'(x)\).

• Differentiability implies continuity: If a function is differentiable at any point, it is also continuous there.
• Derivatives represent the slope of the tangent line at any point on the curve.
• On the interval \([a, b]\), \(f\) not only changes but does so in a predictable manner without jumps or breaks.

Interval Analysis

Interval analysis helps us understand what happens within a specific set of points. For our exercise, we focus on the interval \([a, b]\).
This interval encompasses all points from \(a\) to \(b\), including both endpoints. The function \( f \) behaves within this range, which is crucial for applying the Mean Value Theorem.
In our setup, the function \( f \) is differentiable throughout the closed interval, \([a, b]\), and we're specifically interested in a point \( c \) located within the open interval \((a, b)\).

• The closed interval \([a, b]\) includes both boundary points, ensuring the function is well-defined at these endpoints.
• The open interval \((a, b)\) excludes \(a\) and \(b\), focusing on the points strictly in between.
• We use interval analysis to pinpoint where specific characteristics of \( f \) have to occur, such as \( f'(c) < 0 \).

Calculus Proof

A calculus proof methodically verifies the properties of a function using theorems and logical reasoning. In our scenario, we employ the Mean Value Theorem (MVT), which provides a way to connect the average rate of change of a function over an interval with the instantaneous rate of change at a specific point.
The MVT requires:

• The function to be continuous on the closed interval \([a, b]\).
• The function to be differentiable on the open interval \((a, b)\).

Given these conditions, the MVT asserts that there exists at least one point \( c \in (a, b) \) where: \[f'(c) = \frac{f(b) - f(a)}{b - a}\]For our exercise, we know \( f(b) < f(a) \), implying that the slope \( \frac{f(b) - f(a)}{b - a} \) is negative. Thus, according to the MVT, there is a point \( c \) such that \( f'(c) \) matches this negative slope.

• Applying MVT clarifies relations between overall and instantaneous rates of change.
• Negative slopes indicate that the function decreases over the interval.
• Proofs in calculus ensure the logical consistency and validity of mathematical assertions.