Problem 13 - College Algebra with Corequisite Support

Question

For the following exercises, state the domain and the vertical asymptote of the function. \(f(x)=\log (3 x+1)\)

Step-by-Step Solution

Answer

Domain: \((-\frac{1}{3}, \infty)\); Vertical Asymptote: \(x = -\frac{1}{3}\).

Step 1: Identify the Domain of the Logarithmic Function

The domain of a logarithmic function, such as \(f(x) = \log(3x + 1)\), is determined by the condition that the argument of the logarithm must be greater than zero. This is because the logarithm is only defined for positive numbers. Therefore, we must solve the inequality \(3x + 1 > 0\).

Step 2: Solve the Inequality

To solve the inequality \(3x + 1 > 0\), subtract 1 from both sides to get \(3x > -1\). Then, divide both sides by 3 to isolate \(x\), yielding \(x > -\frac{1}{3}\). This inequality describes the domain of the function.

Step 3: State the Domain

The domain of the function \(f(x) = \log(3x + 1)\) is all real numbers \(x\) such that \(x > -\frac{1}{3}\). In interval notation, this is \((-\frac{1}{3}, \infty)\).

Step 4: Determine the Vertical Asymptote

Vertical asymptotes occur where the argument of the logarithm equals zero, leading to the expression becoming undefined. Set \(3x + 1 = 0\) and solve for \(x\) to find the vertical asymptote.

Step 5: Solve for the Asymptote

Solving \(3x + 1 = 0\), subtract 1 from both sides to get \(3x = -1\). Then, divide by 3 to find \(x = -\frac{1}{3}\). This represents the location of the vertical asymptote.

Step 6: State the Vertical Asymptote

The vertical asymptote of the function is at \(x = -\frac{1}{3}\), which is where the function becomes undefined.

Key Concepts

Vertical Asymptote

A vertical asymptote is a line where a graph incidentally shoots off to infinity at a certain point. It's where a function ceases to be defined, typically indicating a boundary that the function cannot cross. For logarithmic functions, a vertical asymptote occurs where the argument of the logarithm equals zero from the equation. This causes the function to be undefined as the logarithm of zero is not defined. In the function given, \( f(x) = \log(3x + 1) \), the vertical asymptote is found by setting the inside of the log to zero: \( 3x + 1 = 0 \). Solving this gives us \( x = -\frac{1}{3} \). At this point, the graph of \( f(x) \) will approach but never touch or cross \( x = -\frac{1}{3} \). By understanding where this vertical line falls, you can accurately predict how the function behaves near these values, helping to guide interpretations of the graph.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions and have a unique set of characteristics and rules. They are written in the form \( f(x) = \log_b(x) \), where \( b \) is a positive real number, called the base. In this exercise, the function is \( f(x) = \log(3x + 1) \), implicitly using base 10 logarithm by default. For logarithmic functions:

• The function is only defined for positive input values.
• They have a vertical asymptote where the argument equals zero.
• They pass through the point \((1, 0)\) if \( f(x) = \log_b(x) \).

To find the domain of a function like this, ensure the inner expression \( 3x + 1 \) is greater than zero. Solving this leads us to understand that \( x > -\frac{1}{3} \). This restriction assures that the input to the log function maintains positivity, ensuring it operates within its defined course. Recognizing these unique characteristics can greatly help in graphing and solving these functions effectively.

Inequality Solving

Solving inequalities is crucial for defining domains, especially for logarithmic functions where only portions of the number line are valid. In our case, we solve \( 3x + 1 > 0 \) to find the domain of \( f(x) = \log(3x + 1) \). The goal is to isolate \( x \) using basic algebraic manipulations:

• Start with the inequality \( 3x + 1 > 0 \).
• Subtract 1 from both sides, resulting in \( 3x > -1 \).
• Divide each side by 3, leading to \( x > -\frac{1}{3} \).

The solution \( x > -\frac{1}{3} \) signifies that any \( x \) greater than \( -\frac{1}{3} \) will keep the logarithm's input positive, adhering to the function's domain constraints. This example illustrates the importance of step-by-step operations to create valid expressions and comprehend the limitations or boundaries for mathematical functions.