Problem 72 - Calculus: Early Transcendentals

Question

(a) Show that the absolute value function \( F(x) = | x | \) is continuous everywhere. (b) Prove that if \( f \) is a continuous function on an interval, then so is \( | f | \). (c) Is the converse of the statement in part (b) also true? In other words, if \( | f | \) is continuous, does it follow that \( f \) is continuous? If so, prove it. If not, find a counterexample.

Step-by-Step Solution

Answer

(a) The function \(|x|\) is continuous everywhere. (b) \(|f|\) is continuous if \(f\) is continuous. (c) The converse is false; counterexample: \(f(x)=x\) except \(f(0)=1\).

Step 1: Absolute Value Function Definition

The absolute value function is defined as follows: \[ F(x) = |x| = \begin{cases} x, & \text{if } x \geq 0 \ -x, & \text{if } x < 0 \end{cases} \] We need to show that this function is continuous for all real numbers.

Step 2: Continuity of Absolute Value Function at x = 0

To show that \( F(x) \) is continuous at \( x = 0 \), we check the limit:\[ \lim_{x \to 0} |x| = |0| = 0. \]Since the limit and the value at that point are equal, \( F(x) \) is continuous at \( x = 0 \).

Step 3: Continuity of Absolute Value Function for x > 0 and x < 0

For \( x > 0 \), \( F(x) = x \), which is a linear and hence continuous function. For \( x < 0 \), \( F(x) = -x \), also linear and continuous. Therefore, \( |x| \) is continuous for all real numbers \( x eq 0 \) as well.

Step 4: Result of Part (a)

We have shown the absolute value function \( F(x) = |x| \) is continuous at all points: \( x = 0 \), as well as for \( x > 0 \) and \( x < 0 \). Therefore, \( F(x) = |x| \) is continuous everywhere.

Step 5: Assume Function f is Continuous

Let \( f \) be a continuous function on an interval \( I \). We need to prove that \( |f| \) is also continuous on \( I \).

Step 6: Prove |f| is Continuous

Assume \( x_0 \in I \). Since \( f \) is continuous, \( \lim_{x \to x_0} f(x) = f(x_0) \). By properties of limits and the absolute value, \[ \lim_{x \to x_0} |f(x)| = |\lim_{x \to x_0} f(x)| = |f(x_0)|. \] Thus, \( |f| \) is continuous at \( x_0 \), proving \( |f| \) is continuous on \( I \).

Step 7: Analyze the Converse Statement

Now, we consider whether the converse is true: Does the continuity of \( |f| \) imply the continuity of \( f \)?

Step 8: Counterexample for Converse

Consider the function \( f(x) = \begin{cases} x, & x eq 0 \ 1, & x = 0 \end{cases} \). Here, \( |f(x)| = |x| \), which is continuous, but \( f(x) \) is not continuous at \( x = 0 \). This serves as a counterexample, proving the converse is false.

Key Concepts

Absolute Value Function

The absolute value function, denoted as \( F(x) = |x| \), is a simple yet crucial concept in mathematics. It determines the non-negative magnitude of any real number \( x \). The function can be expressed as a piecewise form:

• \( F(x) = x \) for \( x \geq 0 \)
• \( F(x) = -x \) for \( x < 0 \)

This piecewise representation ensures that \( F(x) \) yields a positive output, regardless of whether \( x \) itself is positive or negative.

The continuity of the absolute value function across its domain can be understood by checking its continuity at critical points such as \( x = 0 \). At this point, the function transitions from \( x \) to \(-x\). By taking the limit as \( x \) approaches zero, we find \( \lim_{x \to 0} |x| = 0 \), which matches \( |0| = 0 \). Since the value of the function and its limit coincide at this point, the function is continuous here. By similar reasoning, the function remains continuous for \( x > 0 \) and \( x < 0 \) because they are simply linear functions, which are inherently continuous.

Piecewise Functions

Piecewise functions are defined using different expressions over different intervals of their domain. This flexibility allows complex behaviors to be formulated in a concise manner, such as the absolute value function.

• They can be used to model functions with differing behaviors in different ranges.
• Continuity in piecewise functions demands that at each breakpoint, the limits from all pieces equal the function's value at that point.

Accurate handling of transitions between intervals is critical for completeness. For example, the absolute value function handles positive and negative values with two linear pieces that connect seamlessly at zero, making it continuous.

Practically, a piecewise function like this might appear in real-world situations, such as calculating tax based on income brackets, where different tax rates apply to different income intervals.

Properties of Limits

Understanding the properties of limits is essential for discussing the continuity of functions. Limits help in predicting the behavior of a function as its input approaches some value.

• The limit property that \( \lim_{x \to a} |f(x)| = |\lim_{x \to a} f(x)| \) is instrumental when dealing with absolute values.
• Continuity is often confirmed when the limit of a function as \( x \) approaches any point \( a \) is equal to the function’s value at \( a \).

These properties simplify the analysis of complex functions, including those expressed in piecewise form, by breaking them down into manageable parts. For instance, determining \( \lim_{x \to 0} |x| = 0 \) confirms the absolute value function is continuous, as it aligns perfectly with \( F(0) = 0 \). Therefore, integrating limits into the study of continuity provides a robust mathematical toolkit.

Counterexamples in Mathematics

Counterexamples are pivotal in mathematics for disproving universal claims. They offer essential insights into the boundaries of mathematical theorems and conjectures.

• A counterexample for the converse statement "If \(|f|\) is continuous, then \(f\) is continuous" is crucial.
• Consider \( f(x) = \begin{cases} x, & x e 0 \ 1, & x = 0 \end{cases} \). Here, \(|f(x)| = |x|\) is continuous, but \(f(x)\) is not because it "jumps" at \(x = 0\).

Here, \( f(x) \) being continuous would demand that \( \lim_{x \to 0} f(x) = f(0) \). However, as \( x \to 0 \), \( f(x) \to 0 \), not 1, creating a discontinuity at \( x = 0 \). This counterexample effectively shows that continuity in absolute value does not ensure continuity of the original function, thus disproving the converse of our earlier statement.