Problem 23 - College Algebra with Corequisite Support

Question

For the following exercises, state the domain, range, and \(x\) - and \(y\) -intercepts, if they exist. If they do not exist, write DNE. \(g(x)=\ln (-x)-2\)

Step-by-Step Solution

Answer

Domain: \((-\infty, 0)\); Range: \((-\infty, \infty)\); \(x\)-intercept: \((-e^2, 0)\); \(y\)-intercept: DNE.

Step 1: Determine the Domain

For the function \( g(x) = \ln(-x) - 2 \), we need to find where \( \ln(-x) \) is defined. The natural logarithm function, \( \ln(u) \), is only defined for \( u > 0 \). Therefore, for \(-x > 0 \), it implies that \( x < 0 \). Thus, the domain of \( g(x) \) is all real numbers less than zero, which can be written as \( (-\infty, 0) \).

Step 2: Determine the Range

The range of a natural logarithm function \( \ln(x) \) is all real numbers \((-fty, fty)\), so \( \ln(-x) \) will also have the same range. Subtracting 2 from \( \ln(-x) \) shifts the range vertically downward by 2, but does not change the fact that \(-\infty\) to \(\infty\) covers all values. Thus, the range of \( g(x) \) is also all real numbers \((-fty, fty)\).

Step 3: Find the x-intercept

The \( x \)-intercept occurs when \( g(x) = 0 \). Setting \( g(x) = \ln(-x) - 2 = 0 \), we solve \( \ln(-x) = 2 \). Exponentiating both sides gives \(-x = e^2 \) or \( x = -e^2 \). Therefore, the \( x \)-intercept is \((-e^2, 0)\).

Step 4: Find the y-intercept

The \( y \)-intercept occurs when \( x=0 \), but according to the domain of \( g(x) \), \( x = 0 \) is not included. Thus, the function has no \( y \)-intercept. Hence, the \( y \)-intercept is DNE (Does Not Exist).

Key Concepts

Understanding Domain and Range

In college algebra, determining the domain and range of a function is fundamental. Here, the domain represents all possible input values, or the values of \( x \), that a function can accept. For the logarithmic function \( g(x) = \ln(-x) - 2 \), the domain is all real numbers less than zero. This is because the natural logarithm function, \( \ln(x) \), is only defined for positive values. Thus, \( \ln(-x) \) requires \(-x > 0\) which simplifies to \(x < 0\). Consequently, the domain for this function is \((-\infty, 0)\).

The range of a function, on the other hand, is all the possible output values the function can produce. Since the natural logarithm \( \ln(x) \) can result in any real number from \( -\infty \) to \( \infty \), so can \( \ln(-x) \). Shifting the entire output downward by 2 (because of the \(-2\)) still results in every negative and positive number being possible outputs. Therefore, the range of \( g(x) \) is also all real numbers \((-\infty, \infty)\). By understanding the domain and range, you can better grasp how functions behave and what types of values they can process.

Delving Into Logarithmic Functions

Logarithmic functions are a vital part of college algebra. They are the inverse processes of exponential functions. In simpler terms, if exponentials "grow quickly," logarithms "grow slowly." The natural logarithm, denoted as \( \ln(x) \), is specifically helpful as it uses the constant \( e \) (approximately 2.718) as its base.

With \( g(x) = \ln(-x) - 2 \), we are working with an adjusted natural logarithm. Here, the term \(-x\) signifies that only negative values of \( x \) are considered, which contrasts from the usual \( \ln(x) \) functions that work with positive values. Logarithmic transformations like this one can seem complex but are essential for solving equations, particularly when dealing with growth and decay models commonly seen in sciences and engineering.

• Functions like \( \ln(x) \) are undefined for \( x \leq 0 \), hence the domain restriction.
• Logarithms convert multiplication into addition, simplifying complex equations.
• Remember base-changing properties like \( \log_b(x) = \frac{\ln(x)}{\ln(b)} \) for different bases.

Understanding these basics of logarithms aids in tackling more sophisticated mathematical challenges.

Exploring Intercepts in Functions

Intercepts are points where a graph intersects the axes on a coordinate plane. They help in visualizing and predicting the behavior of functions. The \( x \)-intercept happens where the function crosses the \( x \)-axis. Hence, it's a point where the output \( y \) of the function is zero. For \( g(x) = \ln(-x) - 2 \), setting the equation to zero yields \( \ln(-x) = 2 \). Solving this gives \( x = -e^2 \) for the \( x \)-intercept, or \((-e^2, 0)\) in coordinate terms.

The \( y \)-intercept exists where the function crosses the \( y \)-axis. This occurs at \( x = 0 \). However, for the function \( g(x) \), since the domain is \( x < 0 \), \( x = 0 \) is not within the domain. Therefore, the \( y \)-intercept does not exist for this function. Recognizing intercepts is crucial in analyzing how a function behaves across different regions of a graph.