Problem 67 - Thomas Calculus

Question

Assume that \(f\) is differentiable on \(a \leq x \leq b\) and that \(f(b) < f(a) .\) Show that \(f^{\prime}\) is negative at some point between \(a\) and \(b.\)

Step-by-Step Solution

Answer

By the Mean Value Theorem, there exists a point \(c\) in \((a, b)\) where \(f'(c) < 0\).

Step 1: Understand the Problem

We are given a function \(f\) that is differentiable on the interval \([a, b]\) with \(f(b) < f(a)\) and need to show that \(f'\) (the derivative of \(f\)) is negative at some point between \(a\) and \(b\).

Step 2: Apply the Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function \(f\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists some point \(c\) in \((a, b)\) such that \[ f'(c) = \frac{f(b) - f(a)}{b - a} . \]

Step 3: Calculate the Slope Using MVT

Since we know \(f(b) < f(a)\), it follows that \(f(b) - f(a) < 0\). Therefore, according to the MVT,\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]This implies that the average rate of change is negative. Since \(b - a > 0\), we have \[ f'(c) = \frac{f(b) - f(a)}{b - a} < 0. \]

Step 4: Conclusion

Therefore, by the MVT, there exists at least one point \(c\) in the interval \((a, b)\) such that the derivative \(f'(c) < 0\). This satisfies the condition we needed to prove: that \(f'\) is negative at some point between \(a\) and \(b\).

Key Concepts

Differentiable Functions

A differentiable function is one that has a derivative at every point on its domain. In simpler terms, these functions are smooth and continuous, without any sharp corners or breaks. Differentiability is important because it ensures that we can find the slope of the function at any given point.
For a function to be differentiable on an interval, it must be both continuous and smooth throughout the interval. This means that there are no sudden jumps, holes, or vertical tangents in the function's graph.
The Mean Value Theorem, which is used in the original exercise, relies on the function being differentiable. This requirement allows us to confidently say there is a point where the derivative (or slope) behaves according to specific rules.

Derivative

The derivative of a function is a measure of how the function's output value changes as its input value changes. Essentially, it tells us the rate of change or the slope of the function at a given point.
Mathematically, if we have a function \( f(x) \), the derivative, denoted as \( f'(x) \), is found by taking the limit of the average rate of change as the interval approaches zero. More formally, this is expressed as:

• \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

Knowing the derivative is crucial, as it provides insight into the function's behavior, such as determining where it is increasing or decreasing. In the exercise, finding a negative derivative between points \( a \) and \( b \) indicates that the function is decreasing at some point on this interval.

Intervals

In mathematics, an interval is a set of numbers that includes all values between two endpoints. Intervals are often discussed in the context of functions to specify where certain properties, like differentiability, are valid.

• Closed interval [a, b]: Includes both endpoints \( a \) and \( b \)
• Open interval (a, b): Excludes both endpoints
• Half-open interval \([a, b)\) or \((a, b]\): Includes one endpoint but not the other

In the context of the original exercise, the function is given as differentiable on \([a, b]\), meaning it is smooth and continuous over the entire interval, inclusive of the endpoints. However, for differentiability, the open interval (a, b) is often focused on, especially when applying the Mean Value Theorem. This is because the theorem specifically applies to the behavior inside the interval, excluding the endpoints, where the real 'action' of checking derivatives and slopes occurs.