Problem 63 - Essential Calculus Early Transcendentals

Question

Show that the function \(g(x)=x|x|\) has an inflection point at \((0,0)\) but \(g^{\prime \prime}(0)\) does not exist.

Step-by-Step Solution

Answer

The inflection point is at (0,0), but g''(0) does not exist.

Step 1: Understanding the Function

First, we need to recognize that the function \( g(x) = x|x| \) can be expressed piecewise. For \( x \geq 0 \), \( |x| = x \), so \( g(x) = x \cdot x = x^2 \). For \( x < 0 \), \( |x| = -x \), so \( g(x) = x \times (-x) = -x^2 \). Therefore, \( g(x) \) can be written as \( g(x) = \begin{cases} x^2 & \text{if } x \geq 0 \ -x^2 & \text{if } x < 0 \end{cases} \).

Step 2: Finding the First Derivative

To find where the inflection point might exist, we first calculate the first derivative \( g'(x) \). For \( x \geq 0 \), \( g'(x) = \frac{d}{dx}(x^2) = 2x \). For \( x < 0 \), \( g'(x) = \frac{d}{dx}(-x^2) = -2x \). Thus, \( g'(x) = \begin{cases} 2x & \text{if } x \geq 0 \ -2x & \text{if } x < 0 \end{cases} \).

Step 3: Evaluating the First Derivative at Zero

We check the first derivative at \( x = 0 \). Since the function is piecewise, \( g'(0) \) must be evaluated from the right and the left limit. As both limits tend towards zero, \( g'(0) = 0 \), indicating continuous first derivative at \( x = 0 \).

Step 4: Finding the Second Derivative

Next, find the second derivative to check for an inflection point. Compute \( g''(x) \) piecewise: \( g''(x) = \frac{d}{dx}(2x) = 2 \) for \( x > 0 \) and \( g''(x) = \frac{d}{dx}(-2x) = -2 \) for \( x < 0 \). Note, the second derivative is undefined right at \( x = 0 \) as the limit from the right doesn't equal the limit from the left.

Step 5: Checking for Inflection Point

An inflection point occurs where the second derivative changes sign. As \( g''(x) = 2 \) for \( x > 0 \) and \( g''(x) = -2 \) for \( x < 0 \), the sign of the second derivative does change as \( x \) passes through \( 0 \). Thus, there is an inflection point at \( (0, 0) \), even though \( g''(0) \) is undefined.

Key Concepts

Understanding Piecewise Functions

A piecewise function is a function that is defined by different expressions for different intervals of the input variable. In the example of the function \( g(x) = x|x| \), it transforms into two different expressions based on whether the value of \( x \) is non-negative or negative.
For \( x \geq 0 \), the absolute value \( |x| \) equals \( x \), making the expression for \( g(x) \) simply \( x^2 \). For \( x < 0 \), \( |x| = -x \), thus the expression becomes \( -x^2 \).
This gives us a definition of \( g(x) \) as a piecewise function:
\[ g(x) = \begin{cases} x^2, & \text{if } x \geq 0 \ -x^2, & \text{if } x < 0 \end{cases} \]
Understanding piecewise functions helps us explore how the function behaves over different intervals. This is crucial for examining properties like continuity and differentiability at the boundaries of these intervals.

Calculating the First Derivative

The first derivative of a function indicates the rate at which the function's value changes with respect to changes in its input. It is fundamental in finding critical points, where the derivative is zero or undefined, indicating potential maximum, minimum, or points of inflection.
For the piecewise function \( g(x) \), the first derivative must also be calculated piecewise:

• For \( x \geq 0 \), the function is \( g(x) = x^2 \). Thus, the derivative \( g'(x) = 2x \).
• For \( x < 0 \), where \( g(x) = -x^2 \), the derivative is \( g'(x) = -2x \).

So, the derivative \( g'(x) \) is:
\[ g'(x) = \begin{cases} 2x, & \text{if } x \geq 0 \ -2x, & \text{if } x < 0 \end{cases} \]
By evaluating \( g'(x) \) at \( x = 0 \), we find that both limits (approaching from positive and negative \( x \)) yield \( 0 \), thus \( g'(0) = 0 \). This continuity at \( x = 0 \) simplifies our analysis of changes around this point.

Understanding the Second Derivative

The second derivative of a function provides insight into the function's concavity — whether it opens upwards or downwards — and helps identify inflection points where the curvature changes.
Again, with a piecewise function, we calculate separately:

• For \( x > 0 \), \( g''(x) = \frac{d}{dx}(2x) = 2 \).
• For \( x < 0 \), \( g''(x) = \frac{d}{dx}(-2x) = -2 \).

This means:
\[ g''(x) = \begin{cases} 2, & \text{if } x > 0 \ -2, & \text{if } x < 0 \end{cases} \]
At \( x = 0 \), \( g''(x) \) is undefined because the left-hand and right-hand limits do not match. Understanding the behavior of the second derivative around \( x = 0 \) is important, as it shows that the function's concavity changes, confirming the presence of an inflection point, even though \( g''(0) \) itself is undefined.

Undefined Derivatives and Their Implications

An undefined derivative occurs at points where the limit defining the derivative does not exist consistently from both directions. At such points, analyzing the derivative's approach from either side is crucial to understanding the function's behavior.
For the function \( g(x) = x|x| \), the second derivative \( g''(x) \) becomes undefined at \( x = 0 \). This is because:

• For \( x > 0 \), the derivative is \( 2 \).
• For \( x < 0 \), the derivative is \(-2 \).

Since these do not approach the same value as \( x \) approaches \( 0 \), \( g''(0) \) is undefined. Despite this, the change in sign of the second derivative as \( x \) crosses \( 0 \) (from positive to negative or vice versa) indicates an inflection point at \( (0, 0) \).
This illustrates how undefined derivatives, while complicated, can still offer valuable information about the function's graphical and descriptive properties.