Problem 170 - Calculus Volume 1

Question

Show there is no \(c\) such that \(f(1)-f(-1)=f^{\prime}(c)(2) .\) Explain why the Mean Value Theorem does not apply over the interval \([-1,1]\) . $$ f(x)=\lfloor x\rfloor $$ (Hint: This is called the floor function and it is defined so that \(f(x)\) is the largest integer less than or equal to \(x\)).

Step-by-Step Solution

Answer

MVT does not apply because \( f(x) = \lfloor x \rfloor \) is not continuous on \([-1, 1]\).

Step 1: Understand the Problem Statement

We need to check if the Mean Value Theorem (MVT) applies to the function \( f(x) = \lfloor x \rfloor \) over the interval \([-1, 1]\). Specifically, there should be no \( c \) such that \( f(1) - f(-1) = f'(c)(2) \).

Step 2: Recall the Mean Value Theorem

The Mean Value Theorem 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 at least one \( c \) in \( (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Step 3: Calculate \( f(1) - f(-1) \) for \( f(x) = \lfloor x \rfloor \)

Evaluate \( f(1) \) and \( f(-1) \). \( f(1) = \lfloor 1 \rfloor = 1 \), and \( f(-1) = \lfloor -1 \rfloor = -1 \). So, \( f(1) - f(-1) = 1 - (-1) = 2 \).

Step 4: Check Continuity and Differentiability

The function \( f(x) = \lfloor x \rfloor \) is not continuous at integer points (e.g., \(-1, 0, 1\)), as it has jump discontinuities. Moreover, it is non-differentiable at any integer because it is not continuous there.

Step 5: Conclude that MVT Does Not Apply

Since \( f(x) = \lfloor x \rfloor \) is not continuous over the interval \([-1, 1]\) due to the jump at \( x = 0 \), the mean value theorem does not apply, and thus there cannot be a \( c \) satisfying the MVT conditions.

Key Concepts

Continuity

A function is continuous when you can draw its graph without lifting your pencil from the paper. In mathematical terms, a function \( f(x) \) is continuous at a point \( c \) if the limit as you approach \( c \) from both directions equals \( f(c) \). Basically, you want the points to connect smoothly without any breaks or "jumps."

For the floor function \( f(x) = \lfloor x \rfloor \), continuity gets tricky. This function, defined as the greatest integer less than or equal to \( x \), has breaks at every whole number. At these points, the function "jumps" from one integer value to the next.

For instance:

• At \( x = 1 \), the function jumps from 0 to 1.
• Similarly, at \( x = 0 \), it jumps from -1 to 0.

Hence, \( f(x) = \lfloor x \rfloor \) is not continuous at integer points in its domain.

Differentiability

Differentiability is about whether a function has a derivative at a particular point. A function is differentiable at a point if it isn't bending too sharply or having jagged turns. Simply put, it should look smooth and not have any jumps.

The mathematical way of saying this, is that the function should be continuous and have a defined slope—a tangent line at that point. If a function is not continuous at a point, it certainly isn't differentiable there either.

Since the floor function \( f(x) = \lfloor x \rfloor \) has sharp turns and isn't continuous at integer points, it isn't differentiable there. This lack of smoothness means we won’t find a derivative across these places. Thus, the Mean Value Theorem doesn’t apply as this function doesn’t meet the requirement of being differentiable over certain intervals.

Floor Function

The floor function, represented by \( \lfloor x \rfloor \), is quite unique. It rounds down any real number to the largest integer that's less than or equal to that number. For instance:

• \( \lfloor 2.7 \rfloor = 2 \)
• \( \lfloor -1.3 \rfloor = -2 \)

Floor functions have a specific behavior that contributes to their non-continuity. At every integer point there's an abrupt shift as it skips from one integer value down to another.

This nature of the floor function causes it to not be continuous, introducing us to the concept of jump discontinuity, which we will explore further. Remember, while it serves specific purposes in mathematics, the floor function fails some tests for certain theorems, like the Mean Value Theorem, due to its inherent properties.

Non-Differentiability

When discussing whether a function is differentiable, we essentially ask if we can find a smooth tangent line at each point of the function. Due to its jump discontinuities, the floor function \( \lfloor x \rfloor \) is non-differentiable at each integer point.

Imagine trying to draw a smooth tangent line on a rocky path, where each rock represents an integer value of \( x \). The sharp jumps at these integer points make it impossible to define a single tangent line.

To apply the Mean Value Theorem, the function must be both continuous and differentiable on the interval in question. With \( f(x) = \lfloor x \rfloor \), the lack of differentiability at integer values means it doesn't meet these criteria, highlighting why there’s no \( c \) within the interval where the theorem holds.

Jump Discontinuity

A jump discontinuity happens when there's a sudden shift in the function's values. It's like abruptly moving from one stair step to the next without gradual transition.

The floor function \( \lfloor x \rfloor \) exhibits jump discontinuities at each integer point. For example:

• At \( x = 0 \), the function "jumps" from -1 to 0.
• At \( x = 1 \), it suddenly moves from 0 to 1.

These abrupt changes mean you can’t draw the graph continuously through these points, rendering the function non-continuous and thus non-differentiable there.

Since the Mean Value Theorem requires continuity over the entire interval, the jump discontinuities prevent it from being applicable to the floor function over any interval containing such jumps, like from \([-1, 1]\). This illustrates an essential point: for some functions, complex theorems have restrictions based on simple foundational properties.