Problem 42 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 37-42: Sketch the graph of \(f\). $$ f(x)=\ln (e+x) $$

Step-by-Step Solution

Answer

The graph starts at \((-e, -\infty)\) with increasing behavior, passing through \((0, 1)\), and is concave down.

Step 1: Analyze Domain of the Function

The domain of the function is determined by the expression inside the logarithm, which must be positive. Hence, solve the inequality: \( e + x > 0 \). Simplifies to \( x > -e \). Therefore, the domain of \( f(x) \) is \( x > -e \).

Step 2: Identify Critical Points and Behavior

Determine any points of interest such as intercepts or behavior at the domain limits. Substitute \( x = 0 \), \( f(0) = \ln(e+0) = \ln(e) =1 \). No x-intercepts exist as \( f(x) \) equals zero results in an equation without a solution.

Step 3: Derive Function and Identify Increasing/Decreasing Intervals

Compute the derivative of \( f(x) = \ln(e + x) \). Using the chain rule, \( f'(x) = \frac{1}{e+x} \cdot 1 = \frac{1}{e+x} \). The derivative is positive for all \( x > -e \), indicating the function is increasing in its entire domain.

Step 4: Analyze Concavity with the Second Derivative

Find the second derivative: \( f''(x) = -\frac{1}{(e+x)^2} \). The second derivative is negative for \( x > -e \), indicating the function is concave down over its domain.

Step 5: Sketch the Graph

Plot the behavior and points identified: an increasing function starting from just right of \( x = -e \). Pass through \((0,1)\) and aim to capture its general increasing trend and concavity downward. Exponentially approaches positive infinity as \( x \to \infty \).

Key Concepts

Domain Analysis

Understanding the domain of a function is crucial for graph sketching. It tells us where the function is defined and helps identify the boundaries for x-values we can use. For logarithmic functions like \(f(x) = \ln(e + x)\), the expression inside the logarithm must be greater than zero.
This requirement stems from the fact that logarithms of non-positive numbers are undefined. Here, solving the inequality \(e + x > 0\) for \(x\) gives us \(x > -e\).
This means:

• The function is only defined for x-values that are greater than \(-e\).
• Any point where \(x \leq -e\) becomes irrelevant for this function.

Ensure you confirm the domain before moving on to find other key features of the graph.

Critical Points

Critical points are where we find potential maximums, minimums, or points of inflection on a graph. These are found by setting the derivative of the function to zero or when the derivative is undefined.
For \(f(x) = \ln(e + x)\), calculate \(f'(x)\) to identify these critical points. Here, \(f'(x) = \frac{1}{e+x}\), which is never zero or undefined for \(x > -e\).

• As a result, there are no critical points where the function changes direction.
• We can further examine specific points like \(f(0) = 1\) which marks the y-intercept.

This means the primary concern is behavior at these boundary limits, rather than at isolated points.

Increasing and Decreasing Intervals

The first derivative informs us about where a function is increasing or decreasing. When the derivative \(f'(x) > 0\), the function is increasing; if \(f'(x) < 0\), it is decreasing.
With \(f'(x) = \frac{1}{e+x}\), which is always positive for \(x > -e\), we conclude:

• The function is always increasing within the domain \(x > -e\).

This consistent behavior simplifies the analysis as we need not worry about intervals where the function decreases. When graph sketching, plot these increases smoothly from left to right.

Function Concavity

Concavity describes how a function curves, determined by the second derivative \(f''(x)\). If \(f''(x) > 0\), the graph is concave up; if \(f''(x) < 0\), it is concave down.
For our function, \(f''(x) = -\frac{1}{(e+x)^2}\), which is negative for \(x > -e\). This indicates:

• The function remains concave down across its domain.

This shape affects how the increasing trend appears graphically as always turning downward, akin to an upside-down cup between points.