Problem 68 - Thomas Calculus in SI Units

Question

Let \(f\) be a function defined on an interval \([a, b] .\) What conditions could you place on \(f\) to guarantee that $$\min f^{\prime} \leq \frac{f(b)-f(a)}{b-a} \leq \max f^{\prime}$$ where \(\min f^{\prime}\) and \(\max f^{\prime}\) refer to the minimum and maximum values of \(f^{\prime}\) on \([a, b] ?\) Give reasons for your answers.

Step-by-Step Solution

Answer

Ensuring \( f \) is continuous on \([a,b]\) and differentiable on \((a,b)\) ensures the inequality holds by the Mean Value Theorem.

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a point \( c \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). This implies that the average rate of change of the function over the interval equals the instantaneous rate of change at some point within that interval.

Step 2: Define Requirements for Continuity and Differentiability

To apply the Mean Value Theorem, the function \( f \) must be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). These conditions ensure that \( f' \) is defined and finite at some point in the interval, allowing us to relate \( f'(c) \) to \( \frac{f(b) - f(a)}{b - a} \).

Step 3: Relate the Inequality to Derivative Extremes

If the function \( f \) satisfies the conditions of the Mean Value Theorem, then the value \( \frac{f(b) - f(a)}{b - a} \) will be equal to \( f'(c) \) for some \( c \) in \( (a, b) \). Since \( f' \) is continuous on \([a, b]\), it will achieve a minimum and maximum value on this interval. Thus, \( \min f' \leq f'(c) \leq \max f' \).

Step 4: Conclude with Required Conditions

To guarantee the inequality \( \min f' \leq \frac{f(b) - f(a)}{b - a} \leq \max f' \), the function \( f \) must be continuous on \([a, b]\) and differentiable on \((a, b)\). These conditions ensure that there exists at least one point \( c \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \), which lies between the minimum and maximum of \( f' \) over \([a, b]\).

Step 5: Provide Reasoning for the Answer

The Mean Value Theorem assures us that the existance of a point \( c \) within \( (a, b) \) where \( f'(c) \) equals the average rate of change \( \frac{f(b) - f(a)}{b - a} \). This means that the average rate lies between the minimum and maximum of \( f' \) when \( f \) meets the requirements of continuity and differentiability over the stated intervals.

Key Concepts

Continuity

For a function to satisfy the Mean Value Theorem (MVT), it must be continuous on a closed interval \[a, b\]. Continuity means there are no breaks, jumps, or holes within this interval. To understand continuity, imagine you can draw the function's graph from point \(a\) to point \(b\) without lifting your pencil.
Continuous functions:

• Ensure that limits exist for all points within the interval.
• Do not have abrupt changes in their value.

Continuity on \[a, b\] is crucial because if a function has a discontinuity, the MVT does not apply.
For example, a function that suddenly jumps from one value to another would not have a definable rate of change at that point.

Differentiability

Differentiability is about the ability to take a derivative at each point of the interval \((a, b)\). What this really means is that the function must have a tangent at every point within this open interval. In simpler terms, a differentiable function is smooth without any sharp corners or cusps.
Key points about differentiability:

• A function can be continuous but not differentiable (e.g., functions with corners).
• Differentiability on \((a, b)\) is necessary for the existence of derivatives throughout the interval.

For the MVT to apply, the function must be both continuous on \[a, b\] and differentiable on \((a, b)\). If a function is not differentiable at a point, the MVT cannot ensure a corresponding point \(c\) where the instant change matches the average rate of change.

Average Rate of Change

The average rate of change of a function between two points \(a\) and \(b\) describes how much the function value changes, on average, for each unit increase in the input. It is calculated as \(\frac{f(b) - f(a)}{b - a}\), which represents the slope of the secant line connecting \( (a, f(a)) \) to \( (b, f(b)) \) on the graph.
Features of average rate of change:

• Provides an overall picture of change over an interval.
• Does not account for fluctuations at individual points.

In the context of the MVT, the average rate of change is pivotal because it is the value that the instantaneous rate of change (derivative) at some point within the interval will exactly match.

Instantaneous Rate of Change

The instantaneous rate of change at a specific point \(c\) within the interval \(a, b\) is given by the derivative of the function, \(f'(c)\). It measures how quickly the function is changing at that exact point, and is visualized as the slope of the tangent line to the curve at \(c\).
Characteristics of instantaneous rate of change:

• Can vary significantly between different points within the interval.
• Calculated by taking the derivative at a specific point.

The MVT guarantees that if the conditions of continuity on \[a, b\] and differentiability on \((a, b)\) are met, this instantaneous rate of change \(f'(c)\) will equal the average rate of change \(\frac{f(b) - f(a)}{b - a}\), proving that there exists a point \(c\) where these two rates coincide.