Problem 3 - Basic Analysis: Introduction to Real Analysis

Question

Let \(f(x):=|x|^{3}\). Compute \(f^{\prime}(x)\) and \(f^{\prime \prime}(x)\) for all \(x\), but show that \(f^{(3)}(0)\) does not exist.

Step-by-Step Solution

Answer

\( f'(x) \) and \( f''(x) \) are 0 at \( x = 0 \), but \( f'''(0) \) does not exist due to a discontinuity.

Step 1: Understand the Given Function

The function is given by \( f(x) = |x|^3 \). This is an absolute value function raised to the power of three.

Step 2: Split the Function Based on Absolute Value

Since \( f(x) = |x|^3 \), we can split it into two parts: \( f(x) = x^3 \) for \( x \geq 0 \) and \( f(x) = (-x)^3 = -x^3 \) for \( x < 0 \).

Step 3: Differentiate for Positive x

For \( x > 0 \), \( f(x) = x^3 \). The derivative is \( f'(x) = 3x^2 \).

Step 4: Differentiate for Negative x

For \( x < 0 \), \( f(x) = -x^3 \). The derivative is \( f'(x) = -3x^2 \).

Step 5: Check Derivative at x = 0

At \( x = 0 \), the derivative \( f'(x) \) from the right is 0 and from the left is also 0, so \( f'(0) = 0 \). Hence, \( f'(x) \) is continuous at \( x = 0 \).

Step 6: Find Second Derivative for Positive x

For \( x > 0 \), \( f'(x) = 3x^2 \). The second derivative is \( f''(x) = 6x \).

Step 7: Find Second Derivative for Negative x

For \( x < 0 \), \( f'(x) = -3x^2 \). The second derivative is \( f''(x) = -6x \).

Step 8: Check Second Derivative at x = 0

At \( x = 0 \), the second derivative from the right is 0 and from the left is also 0, so \( f''(0) = 0 \). Hence, \( f''(x) \) is continuous at \( x = 0 \).

Step 9: Evaluate Third Derivative for Positive x

For \( x > 0 \), \( f''(x) = 6x \). The third derivative is \( f'''(x) = 6 \).

Step 10: Evaluate Third Derivative for Negative x

For \( x < 0 \), \( f''(x) = -6x \). The third derivative is \( f'''(x) = -6 \).

Step 11: Check if Third Derivative Exists at x = 0

The third derivative from the right is 6 and from the left is -6. Since these limits are not equal, \( f'''(0) \) does not exist.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It's essentially the process of finding the rate at which a function is changing at any given point. This rate of change is represented by the derivative of the function. When you differentiate a function, you're looking for this derivative, which provides a slope or gradient of the function at any point along its curve.
In real analysis, differentiation helps us understand how functions behave locally. It gives insights about features like increasing or decreasing trends, and can even identify points where the function takes a local maximum or minimum.
To differentiate a function formally, you apply derivative rules, such as the power rule for polynomial functions. For example, the derivative of a simple function like \( f(x) = x^n \) is found using the power rule, giving \( f'(x) = n \cdot x^{n-1} \).
When facing functions involving absolute values, such as \( f(x) = |x|^3 \), differentiation becomes slightly more complex because the absolute value affects the function's behavior at points where the input changes sign. This leads us to carefully segment the function into intervals or use piecewise definitions to approach differentiation accurately.

Absolute Value Functions

Functions involving absolute values can offer unique challenges in real analysis. The absolute value of a number \( |x| \) is its non-negative value, meaning it affects how functions behave, especially around zero.
For an absolute value function like \( f(x) = |x|^3 \), understanding its piecewise nature is crucial. This function can be split into two separate equations, based on the sign of \( x \):

• For \( x \geq 0 \), it's simply \( f(x) = x^3 \).
• For \( x < 0 \), it transforms into \( f(x) = -x^3 \).

This splitting facilitates differentiation. Each part of the function is treated as a standard polynomial during this process. It's important to recognize points where the function changes form, especially at \( x=0 \), where the nature of the function transitions smoothly but introduces a point of continuity and differentiability check.
Absolute value functions can be tricky when computing higher-order derivatives, as seen in this exercise. Careful segmentation and attentive analysis at transit points, like zero, help in exactly understanding the behavior of the function.

Derivatives

Derivatives are fundamental to understanding how functions change. They provide the slope or rate of change of a function at any point and are crucial for analyzing many characteristics of functions, such as increasing and decreasing intervals, and points of inflection.
For the function \( f(x) = |x|^3 \), derivatives are calculated separately over positive and negative domains due to the absolute value. The first derivative tells us the slope at any point along the curve:

• For \( x > 0 \), \( f'(x) = 3x^2 \).
• For \( x < 0 \), \( f'(x) = -3x^2 \).

The equality \( f'(0) = 0 \) indicates a continuous derivative at zero, signifying a smooth point.
The second derivative, indicating how the slope of the function itself is changing, is similarly split:

• For \( x > 0 \), \( f''(x) = 6x \).
• For \( x < 0 \), \( f''(x) = -6x \).

Here, too, \( f''(0) = 0 \) confirms continuity.
However, discrepancies arise with the third derivative:

• For \( x > 0 \), \( f'''(x) = 6 \).
• For \( x < 0 \), \( f'''(x) = -6 \).

Since these don't equalize at zero, \( f'''(0) \) doesn't exist, depicting a break in smoothness in the third derivative at that point. This highlights how derivatives help us observe and diagnose crucial changes and anomalies in function behavior.