Problem 48 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test or the Second Derivative Test to determine the relative extreme values, if any, of the function. Then sketch the graph of the function. $$ f(x)=\ln (1+x)+\ln (1-x) $$

Step-by-Step Solution

Answer

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

Step 1: Determine Domain

The function is given by \[ f(x) = \ln(1+x) + \ln(1-x) \]Since logarithm is only defined for positive arguments, \(1+x > 0\) and \(1-x > 0\). Solving these inequalities, we find that \(-1 < x < 1\). Thus, the domain of \(f(x)\) is \((-1, 1)\).

Step 2: Find the First Derivative

Differentiate the function with respect to \(x\):\[ f'(x) = \frac{d}{dx}\left(\ln(1+x) + \ln(1-x)\right) = \frac{1}{1+x} - \frac{1}{1-x} \]This simplifies to:\[ f'(x) = \frac{1 - x}{(1+x)(1-x)} - \frac{1 + x}{(1-x)(1+x)} = \frac{1-x - (1+x)}{(1-x^2)} = \frac{-2x}{1-x^2} \]

Step 3: Find Critical Points

Set \(f'(x) = 0\) to find critical points:\[ \frac{-2x}{1-x^2} = 0 \]The numerator \(-2x\) equals zero when \(x = 0\). Thus, the critical point is \(x = 0\).

Step 4: Use the Second Derivative Test

Find the second derivative to apply the test:\[ f''(x) = \left(\frac{d}{dx}\right)(\frac{-2x}{1-x^2}) \]Using the quotient rule, we get:\[ f''(x) = \frac{(1-x^2)\cdot(-2) - (-2x)\cdot(-2x)}{(1-x^2)^2} = \frac{-2 + 2x^2 - 4x^2}{(1-x^2)^2} = \frac{-2(1 + x^2)}{(1-x^2)^2} \]Evaluate \(f''(x)\) at \(x = 0\):\[ f''(0) = \frac{-2(1 + 0^2)}{(1-0^2)^2} = -2 \]Since \(f''(0) < 0\), \(x = 0\) is a relative maximum.

Step 5: Sketch the Graph

Based on the analysis, the function has a relative maximum at \(x = 0\) with value \(f(0) = 0\). The function is defined on \((-1, 1)\) and decreases towards \(-\infty\) as \(x\) approaches \(-1\) or \(1\). Sketch the graph with these properties in mind, showing a peak at \(x = 0\) and decreasing towards the endpoints of the domain.

Key Concepts

First Derivative Test

The First Derivative Test is a valuable tool in calculus for identifying whether a function has a relative minimum or maximum at a critical point. A critical point occurs where the derivative of the function equals zero or is undefined. To apply the First Derivative Test, you need to examine the sign of the first derivative before and after the critical point.

This test works by observing changes in the sign of the first derivative, denoted as \(f'(x)\). Here's how it works:

• If \(f'(x)\) changes from positive to negative at a critical point, the function changes from increasing to decreasing, indicating a relative maximum.
• If \(f'(x)\) changes from negative to positive, the function changes from decreasing to increasing, indicating a relative minimum.
• If there is no sign change, then the function does not have a relative extremum at that point.

In our example, although the First Derivative Test could be applied, we used the Second Derivative Test for ease. However, understanding this test is crucial for analyzing other functions and cases where the second derivative test might not apply.

Second Derivative Test

The Second Derivative Test helps identify the relative extrema of a function by evaluating the concavity of its graph. This test relies on the second derivative of the function, \(f''(x)\), to provide deeper insights into the behavior of the function near its critical points.

Here's the process for using the Second Derivative Test:

• Find the critical points, where the first derivative \(f'(x) = 0\) or is undefined.
• Calculate the second derivative, \(f''(x)\), and evaluate it at each critical point.
• If \(f''(x) > 0\), the function is concave up at that point, and the critical point is a relative minimum.
• If \(f''(x) < 0\), the function is concave down at that point, and the critical point is a relative maximum.
• If \(f''(x) = 0\), the test is inconclusive, and further analysis is needed.

In the exercise, \(f''(0) = -2\), which is negative. This indicates the function has a relative maximum at \(x=0\). This conclusion aligns neatly with the behavior of the natural log function used, emphasizing the power of the Second Derivative Test.

Critical Points

Critical points are central to identifying the significant features of a function's graph, such as peaks, troughs, or points of inflection. They occur where the first derivative equals zero or is undefined, indicating a potential change in the function’s direction.

Steps to find critical points include:

• Taking the first derivative of the function, \(f'(x)\).
• Setting \(f'(x) = 0\) to find potential critical points.
• Solving for \(x\) to locate these points.

In our example, setting \(f'(x) = \frac{-2x}{1-x^2} = 0\) leads to \(x = 0\) as the sole critical point within the domain \((-1,1)\). This point is key in assessing where the function has either a relative minimum or maximum, as further determined by the Second Derivative Test.

Understanding critical points positions you to delve deeper into the function's graph, achieving a comprehensive view of its dynamic nature.