Problem 46 - Calculus Early Transcendentals

Question

Use any method to find the relative extrema of the function \(f .\) $$ f(x)=\ln \left|2+x^{3}\right| $$

Step-by-Step Solution

Answer

The function \(f(x)\) has a relative maximum at \(x=0\).

Step 1: Find the Derivative

To locate the relative extrema, we first need to find the derivative of the function. The function is given by \(f(x) = \ln|2 + x^3|\). Recall that the derivative of \(\ln|u|\) with respect to \(x\) is \(\frac{u'}{u}\), where \(u = 2 + x^3\). Hence, the derivative is \(f'(x) = \frac{3x^2}{2+x^3}\).

Step 2: Set the Derivative Equal to Zero

Relative extrema occur where the derivative is zero or undefined. Set \(f'(x)=\frac{3x^2}{2+x^3}=0\). This equation is zero when the numerator is zero, leading to \(3x^2=0\), which gives \(x=0\).

Step 3: Check Derivative’s Undefined Points

Check where \(f'(x) = \frac{3x^2}{2+x^3}\) might be undefined. The derivative is undefined when the denominator is zero. Set \(2+x^3=0\), solving for \(x\) gives \(x = -\sqrt[3]{2}\).

Step 4: Determine the Sign Changes

To determine if these points are relative extrema, check the sign changes of \(f'(x)\) around \(x=0\) and \(x=-\sqrt[3]{2}\). For \(x < -\sqrt[3]{2}\), choose a test point in the derivative to see it is positive. For \(-\sqrt[3]{2} < x < 0\), it is negative. For \(x > 0\), it is positive again.

Step 5: Identify the Relative Extrema

Based on the sign changes, there is a relative maximum at \(x=0\) because the derivative goes from negative to positive. \(x=-\sqrt[3]{2}\) is not a relative extremum as the function changes from positive to negative.

Key Concepts

Derivative Calculus

Derivative calculus is a fundamental process in calculus that helps us analyze how functions change. When exploring extrema (highs and lows) of any function, we first turn to its derivative. Essentially, the derivative provides a slope indication – it tells us how steeply a function is increasing or decreasing at any given point. For a function like \(f(x) = \ln|2 + x^3|\), the derivative is computed using the chain rule:

• The chain rule states \( \frac{d}{dx} \ln|u| = \frac{u'}{u}\).
• Here our \(u\) is \(2 + x^3\), and \(u'\) would be \(3x^2\).

This results in the derivative \(f'(x) = \frac{3x^2}{2+x^3}\). Finding this derivative is crucial because it sets the stage for identifying critical points, helping us pinpoint potential relative extrema by showing where the function's slope is zero or undefined.

Critical Points

Critical points occur where the derivative of a function is either zero or undefined. These points mark potential candidates for relative extrema (maximum or minimum points) on the function:

• To find where the derivative is zero, set \(f'(x) = \frac{3x^2}{2+x^3} = 0\). This happens when the numerator, \(3x^2\), is zero, providing \(x = 0\) as a critical point.
• Next, examine where the derivative becomes undefined, which occurs if the denominator, \(2+x^3\), equals zero. Solving \(2+x^3=0\) gives \(x = -\sqrt[3]{2}\) as another point of interest.

By identifying these critical points, we outline potential locations for turning behavior in the function – effectively where it could have peaks or troughs. Once identified, further analysis regarding the direction of change around these points is necessary to confirm if they indeed represent relative extrema.

Sign Change Method

The sign change method helps confirm the nature of critical points by examining the behavior of the derivative around these points. By observing the change in sign of \(f'(x)\), we can determine whether each critical point is a maximum, minimum, or neither:

• Near \(x = 0\), observe the behavior to the left and right. If \(f'(x)\) transitions from negative to positive, it suggests a relative minimum. Conversely, from positive to negative, it indicates a relative maximum. In this case, as \(x\) goes from negative to positive around \(0\), the derivative changes from negative to positive, indicating a maximum at \(x = 0\).
• For \(x = -\sqrt[3]{2}\), our observation shows \(f'(x)\) transitions from positive to negative. However, unless surrounded by an opposite sign (leading to a turning point), it does not mark a relative extremum.

Implementing the sign change method effectively validates or rejects potential extrema by considering the derivative's local behavior, offering insights into whether critical points truly represent peaks or dips on the curve.