Problem 50 - Calculus Early Transcendentals

Question

Use any method to find the relative extrema of the function \(f .\) $$ f(x)=|1+\sqrt[3]{x}| $$

Step-by-Step Solution

Answer

Relative minimum at \( x = -1 \) with \( f(-1) = 0 \).

Step 1: Understand the Function Form

The given function is \( f(x) = |1 + \sqrt[3]{x}| \). This function involves an absolute value and a cube root, meaning we need to consider where the expression inside the absolute value changes sign.

Step 2: Analyze the Expression Inside the Absolute Value

The expression inside the absolute value is \( 1 + \sqrt[3]{x} \). This changes sign when \( \sqrt[3]{x} = -1 \), so create the equation \( 1 + \sqrt[3]{x} = 0 \) which gives \( x = -1 \). Thus, evaluate \( f(x) \) separately on intervals \( (-\infty, -1) \), \( -1 \), and \( (-1, \infty) \).

Step 3: Evaluate \( f(x) \) on Different Intervals

For \( x < -1 \): \( \sqrt[3]{x} < -1 \), so \( 1 + \sqrt[3]{x} < 0 \) and \( f(x) = -(1 + \sqrt[3]{x}) = -1 - \sqrt[3]{x} \).For \( x = -1 \): \( 1 + \sqrt[3]{-1} = 0 \), hence \( f(-1) = 0 \).For \( x > -1 \): \( \sqrt[3]{x} > -1 \), so \( 1 + \sqrt[3]{x} > 0 \) and \( f(x) = 1 + \sqrt[3]{x} \).

Step 4: Determine Critical Points by Derivative Test

Derive \( f(x) \), differentiate inside each interval.For \( x < -1 \): \( f'(x) = -\frac{1}{3x^{2/3}} \) which is negative, indicating a decreasing function.At \( x = -1 \), f value changes, check sign change.For \( x > -1 \): \( f'(x) = \frac{1}{3x^{2/3}} \) which is positive, indicating an increasing function.

Step 5: Identify the Relative Extrema

The point \( x = -1 \) is where the derivative changes sign from negative to positive, indicating a relative minimum, with \( f(-1) = 0 \). No maximum as function unbounded elsewhere.

Key Concepts

Absolute Value Function

The absolute value function is a fundamental mathematical concept that gives the distance of a number from zero on a number line, ignoring any negative sign. In the context of this exercise, the absolute value function

• is applied to the expression inside: \(1 + \sqrt[3]{x}\),
• ensures that \(f(x)\) remains non-negative for all \(x\).

This characteristic is especially important when determining the behavior of the function over different intervals. When the expression inside the absolute value is positive, it remains unchanged. However, when it's negative, it flips sign to become positive. For the given function \(f(x) = |1 + \sqrt[3]{x}|\):

• On the interval \(x > -1\), the expression \(1 + \sqrt[3]{x} > 0\), and thus \(f(x) = 1 + \sqrt[3]{x}\).
• On the interval \(x < -1\), the expression \(1 + \sqrt[3]{x} < 0\), thus \(f(x) = -(1 + \sqrt[3]{x}) = -1 - \sqrt[3]{x}\).

Understanding these behaviors is key to analyzing the rest of the exercise.

Derivative Test

To find relative extrema of a function, the derivative test is frequently used. The derivative of a function provides information about its slope, and changes in the derivative's sign indicate potential relative maxima or minima. In this exercise, we are interested in the derivative of the function \(f(x) = |1 + \sqrt[3]{x}|\), which requires special consideration at the point where the absolute value changes sign, namely, \(x = -1\). Calculate the derivative separately depending on the intervals:

• For \(x < -1\), since \(f(x) = -1 - \sqrt[3]{x}\), the derivative is \(f'(x) = -\frac{1}{3x^{2/3}}\).
• For \(x > -1\), since \(f(x) = 1 + \sqrt[3]{x}\), the derivative is \(f'(x) = \frac{1}{3x^{2/3}}\).

The derivative is not defined exactly at \(x = -1\) because of the abrupt change caused by the absolute value, but the changes in sign around this point help us determine the nature of the extrema.

Sign Change Analysis

Sign change analysis is used to understand how the derivative changes across intervals of the function. This allows us to determine whether the function is increasing or decreasing on those intervals, providing insight into the locations of relative extrema.For the function \(f(x) = |1 + \sqrt[3]{x}|\), we observe:

• When \(x < -1\), the derivative \(f'(x) = -\frac{1}{3x^{2/3}}\) is negative, indicating the function is decreasing.
• When \(x > -1\), the derivative \(f'(x) = \frac{1}{3x^{2/3}}\) is positive, meaning the function is increasing.

The critical point at \(x = -1\) is where the sign of the derivative changes from negative to positive. This shift signifies a relative minimum at \(x = -1\) as the function transitions from decreasing to increasing.

Increasing and Decreasing Functions

Understanding when a function is increasing or decreasing is crucial for identifying relative extrema. This involves analyzing the function's derivative, as the sign of the derivative informs us about the behavior of the original function.For the function \(f(x) = |1 + \sqrt[3]{x}|\), the derivative provides details about the function's rate of change in different intervals:

• For \(x < -1\), since \(f'(x) = -\frac{1}{3x^{2/3}}\) is negative, the function is decreasing in this interval. This means as \(x\) moves towards \(-1\), the value of \(f(x)\) decreases.
• For \(x > -1\), with \(f'(x) = \frac{1}{3x^{2/3}}\) positive, the function is increasing. Here, as \(x\) increases, so does \(f(x)\).

The transition from decreasing (as \(x\) approaches \(-1\) from the left) to increasing (as \(x\) goes beyond \(-1\)) confirms that \(x = -1\) is a relative minimum point for this function.