Problem 24 - Calculus for Biology and Medicine

Question

Suppose that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b) .\) Show that if \(f^{\prime}(x)<0\) for all \(x \in(a, b)\), then \(f\) is decreasing on \([a, b] .\)

Step-by-Step Solution

Answer

If \( f'(x) < 0 \) for all \( x \in (a, b) \), then \( f \) is decreasing on \([a, b]\) by the Mean Value Theorem.

Step 1: Define Decreasing Function

A function \( f \) is said to be decreasing on an interval \( [a, b] \) if for any two numbers \( x_1, x_2 \) within \([a, b]\), whenever \( x_1 < x_2 \), it follows that \( f(x_1) > f(x_2) \). We need to use this definition to show that \( f \) is decreasing on \( [a, b] \) given the condition \( f'(x) < 0 \).

Step 2: Apply Mean Value Theorem

According to the Mean Value Theorem, if a function \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a point \( c \in (a, b) \) such that \[ f'(c) = \frac{f(b) - f(a)}{b - a} . \] Since \( f'(x) < 0 \) for all \( x \in (a, b) \), the slope \( f'(c) \) is negative.

Step 3: Conclude Decreasing Behavior

Given \( f'(c) < 0 \), it follows from the Mean Value Theorem that \( \frac{f(b) - f(a)}{b - a} < 0 \). Hence, \( f(b) - f(a) < 0 \), which implies \( f(b) < f(a) \). Since the function decreases over any sub-interval due to the negative derivative, \( f \) is decreasing on \([a, b]\).

Key Concepts

Mean Value Theorem

The Mean Value Theorem is a fundamental principle in calculus. It applies to functions that are continuous on a closed interval \( [a, b] \) and differentiable on the open interval \( (a, b) \). The theorem guarantees the existence of at least one point \( c \) in the interval \( (a, b) \) where the derivative \( f'(c) \) equals the average rate of change of the function over \( [a, b] \). In formula form: \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]
This theorem is crucial because it links the concept of a function's derivative with its values at the endpoints of an interval. When applied, it provides insight into the behavior of the function between \( a \) and \( b \). If the derivative is negative, as in our exercise, it hints at a decreasing behavior of the function.

Differentiability

Differentiability of a function on an interval means that the function has a derivative at each point within that interval. In other words, if a function \( f \) is differentiable at a point \( x \), the derivative \( f'(x) \) exists at that point. For a function to be differentiable, it must also be continuous, though the reverse is not necessarily true.
Differentiability implies smoothness of the curve of the function, without any sharp turns or cusps, in the subject interval. This concept is essential for the application of the Mean Value Theorem as it assures that the function behaves predictably between points on the defined interval. In our exercise, knowing that the function is differentiable on \( (a, b) \) assures us of the existence of the derivative, guiding us toward proving the decreasing nature of \( f \).

Decreasing Functions

In calculus, a function \( f \) is considered decreasing on an interval \( [a, b] \) if, for every two numbers \( x_1 \) and \( x_2 \) in this interval where \( x_1 < x_2 \), it holds that \( f(x_1) > f(x_2) \). This definition indicates that as you move from left to right across the interval, the values of the function decrease.
This notion is significantly connected with the derivative of the function. If \( f'(x) < 0 \) for all \( x \) in the interval \( (a, b) \), just like in the given problem, it implies that \( f \) is decreasing. Essentially, a negative derivative signifies a downward slope, confirming that the function is decreasing when observed from any point within the interval. This understanding aligns with the conclusion derived via the Mean Value Theorem in the previous problem solution.