Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ f(x)=\ln (1+x) $$

Step-by-Step Solution

Answer

Domain: \( x > -1 \); Increasing and concave down for \( x > -1 \).

Step 1: Understand the given function

The function provided is \( f(x) = \ln(1+x) \). It is a natural logarithmic function where the logarithm is applied to \( 1 + x \).

Step 2: Domain of the function

Since the logarithmic function \( \ln(y) \) is defined only for \( y > 0 \), we need \( 1+x > 0 \). Thus, the domain of the function is \( x > -1 \).

Step 3: First derivative to find the rate of change

To find the first derivative \( f'(x) \) of \( \ln(1+x) \), we apply the chain rule. Since the derivative of \( \ln(u) \) is \( \frac{1}{u} \), we have:\[ f'(x) = \frac{1}{1+x} \cdot 1 = \frac{1}{1+x} \]

Step 4: Critical points and intervals of increase/decrease

The derivative tells us the rate of change of the function. Since \( \frac{1}{1+x} \) is positive for \( x > -1 \), the function is increasing in its entire domain \( x > -1 \).

Step 5: Second derivative to analyze concavity

Calculate the second derivative \( f''(x) \) to understand concavity:\[ f''(x) = \frac{d}{dx} (\frac{1}{1+x}) = -\frac{1}{(1+x)^2} \]This results in a negative value for all \( x > -1 \), indicating the function is concave downward across its domain.

Step 6: Sketch and behavior analysis

The function \( \ln(1+x) \) is defined for \( x > -1 \), increasing over its entire domain, and concave down. It approaches infinity as \( x \) approaches infinity and has a vertical asymptote at \( x = -1 \).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (usually represented by 'x') that will result in a valid output for the function. For natural logarithmic functions, the function is only defined when the argument (the part inside the logarithm) is greater than zero. In our given function, \[ f(x) = \ln(1+x) \]the argument is \[ 1 + x \]. To find the domain, we solve the inequality: \[ 1 + x > 0 \]. Thus, \[ x > -1 \]. This means the domain includes all x-values greater than -1, excluding x = -1 itself since it would make the argument equal to zero, which is undefined in logarithms.

• Remember: Logarithmic functions require a positive argument.
• Check the argument to determine the domain.

Identifying the domain is crucial as it tells us the range within which the function can safely be evaluated.

Derivative

The derivative of a function gives us important information about the rate at which the function's values change with respect to changes in its input. For the natural logarithmic function \( f(x) = \ln(1+x) \), the first derivative \( f'(x) \) tells us how the function is increasing or decreasing.
In this case, the derivative is calculated using the chain rule, and for our function, it simplifies to: \[ f'(x) = \frac{1}{1+x} \]. This derivative is positive when \( x > -1 \) because the denominator \( 1+x \) is positive, ensuring that \( f(x) \) is increasing for all x in its domain:

• The function is always increasing as x increases.

Understanding the derivative is essential, especially when analyzing the behavior of the function such as identifying intervals of increase or decrease and locating critical points.

Concavity

Concavity describes the direction in which a function curves. To determine this for \( f(x) = \ln(1+x) \), we calculate the second derivative.
Computing the second derivative, we find: \[ f''(x) = -\frac{1}{(1+x)^2} \]. Since this expression is negative for all \( x > -1 \), it tells us that the function is concave downward in its entire domain:

• Concave downward: The curve bends downwards.

Visually, a concave downward shape appears like an upside-down bowl across the domain. This behavior can give insights on the potential behavior of functions, for example, indicating that its rate of increase is slowing down as x gets larger.