Problem 12 - College Algebra with Corequisite Support

Question

For the following exercises, state the domain and the vertical asymptote of the function. \(g(x)=\ln (3-x)\)

Step-by-Step Solution

Answer

Domain: \((-\infty, 3)\); Vertical Asymptote: \(x = 3\).

Step 1: Understand the Function

The function given is a logarithmic function, specifically the natural logarithm form, \(g(x) = \ln(3-x)\). Logarithms are only defined for positive arguments, so we must find where \(3-x > 0\).

Step 2: Inequality for Domain

To find the domain, solve the inequality \(3-x > 0\). This means rearranging the inequality.

Step 3: Solve the Inequality

Rearrange the inequality \(3-x > 0\) to find \(x < 3\). This indicates that the function is defined for all values of \(x\) less than 3.

Step 4: State the Domain

The function \(g(x) = \ln(3-x)\) is defined on the interval \((-\infty, 3)\), meaning all real numbers less than 3.

Step 5: Identify the Vertical Asymptote

A vertical asymptote occurs where the argument of the logarithm function approaches 0. As \(x\) approaches 3 from the left, \(3-x\) approaches 0. Hence, \(x = 3\) is the location of the vertical asymptote.

Step 6: Conclusion

The domain of the function is the set of real numbers less than 3, \((-\infty, 3)\), and there is a vertical asymptote at \(x = 3\).

Key Concepts

Logarithmic Functions

Logarithmic functions are fundamental to mathematics, and specifically in calculus, they help us model various phenomena such as growth and decay. A logarithmic function often appears in the form of \(f(x) = \ln(x - a)\), where \(a\) shifts the graph horizontally. In the function \(g(x) = \ln(3-x)\), the argument \((3-x)\) plays a crucial role in determining the domain of the function because logarithms are only defined for positive arguments. This means the more immediate the argument goes to zero, the more crucial it becomes to determine the set values for which \(3-x\) remains positive.

Inequalities

Understanding inequalities is key to solving problems regarding the domain of logarithmic functions. Here, we have the condition \(3-x > 0\). To find out which values \(x\) can take, we solve that inequality.

• Subtract \(-x\) from both sides to have \(-x > -3\).


• Multiply every side by \(-1\), and remember to change the inequality direction. This gives us \(x < 3\).

Inequalities like these help delineate the parts of the graph where the function exists. By solving \(3-x > 0\), we find that \(x\) must be less than 3 for the function to remain defined.

Function Behavior

Function behavior, especially around critical points, gives us insight into how a function works. Here, the critical point is where \(x = 3\) and it plays a prominent role because as \(x\) approaches 3 from the left, \(3-x\) gets closer to zero yet remains positive. This approaching zero behavior indicates that the logarithmic function will approach negative infinity. The graph therefore has a vertical asymptote at \(x = 3\). Observing this, it's understood that the function displays a dramatic change around this point but remains continuous for all \(x < 3\).

Calculus Concepts

Calculus provides tools for analyzing the behavior of functions, particularly around critical points like vertical asymptotes. In this case, since \(g(x) = \ln(3-x)\) approaches negative infinity as \(x\) nears 3 from the left, the vertical asymptote at \(x = 3\) signifies an essential discontinuity. It reflects a boundary for the domain of the function.

• As \(x \to 3^-\), \(g(x) \to -\infty\), illustrating rapid decline.

This understanding underscores that logarithmic functions in calculus aren't just static—they exhibit dynamic changes, heavily influenced by asymptotic boundaries. Such boundaries are crucial in both theoretical and practical mathematical analyses.