Question

Show that \(f^{\prime}\) need not be continuous or bounded on \([a, b]\) (under the standard metric), even if \(f\) is differentiable there.

Short answer

The function \( f(x) = x^2 \sin(1/x) \) is differentiable, but \( f' \) is neither continuous nor bounded.

Step-by-step solution

Understand the Problem

The problem asks us to find an example of a differentiable function \( f \) on a closed interval \([a, b]\) such that its derivative \( f' \) is not continuous or bounded.

Choose a Function

Choose a differentiable function that has a derivative which is not continuous. A classic example is the function \( f(x) = x^2 \sin(1/x) \) for \( x eq 0 \) and \( f(0) = 0 \) on the interval \([-1, 1]\).

Compute the Derivative

Calculate the derivative of the function \( f(x) = x^2 \sin(1/x) \) using the product and chain rule: \[ f'(x) = 2x \sin(\frac{1}{x}) - \cos(\frac{1}{x}) \]. Note that for \( x = 0 \), you need to use definition of derivative to show that \( f'(0) = 0 \).

Analyze Continuity of the Derivative

Examine where \( f'(x) \) is discontinuous. Notice how the term \(- \cos(\frac{1}{x})\) heavily oscillates as \( x \to 0 \), making \( f'(x) \) discontinuous at \( x = 0 \).

Check if \( f' \) is Bounded

Search for bounds of \( f'(x) \). Observe that, due to the term \(-\cos(\frac{1/x})\), \( f'(x) \) oscillates between \(-1\) and \(1\) infinitely as \( x \to 0 \). Therefore, \( f'(x) \) is not bounded as \( x \to 0 \).

Conclude the Example

The derivative \( f' \) is discontinuous at \( x = 0 \) and not bounded on \([-1, 1]\). Thus, even though \( f \) is differentiable everywhere on \([-1, 1]\), \( f' \) can be neither bounded nor continuous.

Key concepts

Continuous Functions

Continuous functions are fundamental in calculus, describing functions without any 'jumps' or 'breaks' in their graphs. Mathematically, a function \( f(x) \) is continuous at a point \( x = c \) if the limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \). That is:\[ \lim_{{x \to c}} f(x) = f(c). \]This means you can draw its graph without lifting your pencil. Continuous functions are nice to work with because they behave predictably.
In the context of differentiability, if a function is differentiable at a point, it is also continuous at that point. However, the reverse is not always true. A function can be continuous everywhere but not differentiable at some points. Beyond individual points, we often study functions that are continuous across intervals, such as \([a, b]\). In these cases, the function does not break or have infinite oscillations within the interval.

Bounded Functions

A function is considered bounded on an interval if there exists real numbers \( M \) and \( N \) such that every value of the function on that interval lies between these two bounds. More formally, a function \( f(x) \) is bounded on an interval \([a, b]\) if:\[ M \leq f(x) \leq N \quad \text{for all} \quad x \in [a, b]. \]This implies that the function doesn't "blow up" to infinity within that range. Bounded functions are particularly important in mathematical analysis because they ensure that a function’s output remains controlled, which is especially useful when analyzing real-world problems.
In differentiability, when discussing the boundedness of a derivative \( f'(x) \), we look to see if its values are capped within some range on a given interval. However, as the example function in the exercise shows, a differentiable function can have a derivative that "escapes" those bounds — meaning it oscillates infinitely or approaches infinity — making it unbounded, despite the original function being neatly defined and differentiable over the entire interval.

Derivative Discontinuity

While a function can be differentiable across an interval, its derivative might not be continuous. This phenomenon is particularly interesting when studying the behavior of the derivative itself. Let's explore this concept further.Consider a function \( f(x) = x^2 \sin(1/x) \) when \( x eq 0 \) and \( f(0) = 0 \). This function is differentiable everywhere, yet its derivative, \( f'(x) = 2x \sin(\frac{1}{x}) - \cos(\frac{1}{x}) \), is a prime example of discontinuity.
As \( x \) approaches 0, the term \(-\cos(\frac{1}{x})\) in the derivative causes heavy and rapid oscillations. These oscillations prevent \( f'(x) \) from having a well-defined limit as \( x \to 0 \). Thus, \( f'(x) \) is discontinuous at \( x = 0 \). This illustrates that a derivative's continuity isn't guaranteed even when the function itself is well-behaved, i.e., differentiable everywhere. Understanding where derivatives can be discontinuous helps in grasping subtler behavioral aspects of calculus problems.