Problem 13 - College Algebra

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: Understanding Logarithm Domain

The domain of the function depends on where the logarithm's argument is defined. A logarithm, such as \(\log(x)\), is defined for positive values of \(x\). Therefore, for \(\log(3x + 1)\), the argument \(3x + 1\) must be greater than zero.

Step 2: Solving for Domain

To find where the expression \(3x + 1\) is greater than zero, we solve the inequality:\[ 3x + 1 > 0 \]Subtract 1 from both sides:\[ 3x > -1 \]Divide both sides by 3:\[ x > -\frac{1}{3} \]Therefore, the domain of \(f(x)\) is \(( -\frac{1}{3}, \infty)\).

Step 3: Understanding Vertical Asymptote

A vertical asymptote in a logarithmic function occurs where its argument approaches zero. Since the argument \(3x + 1\) of our function goes to zero when \(3x + 1 = 0\), we solve this equation to find the vertical asymptote.

Step 4: Solving for Vertical Asymptote

Set the argument \(3x + 1\) equal to zero and solve for \(x\):\[ 3x + 1 = 0 \]Subtract 1 from both sides:\[ 3x = -1 \]Divide by 3:\[ x = -\frac{1}{3} \]Therefore, the vertical asymptote is at \(x = -\frac{1}{3}\).

Key Concepts

Vertical Asymptote

When discussing functions, the term "vertical asymptote" often comes up. A vertical asymptote is a line that a graph of a function approaches, but never actually touches or crosses. In simpler terms, it represents a value that the independent variable (usually denoted as \(x\)) can get infinitely close to but never reach. In logarithmic functions, vertical asymptotes occur where the argument of the logarithm equals zero.
For the function \(f(x) = \log(3x + 1)\), the vertical asymptote happens where the inside of the logarithm, \(3x + 1\), equals zero. Solving \(3x + 1 = 0\) shows us this critical point, where \(x = -\frac{1}{3}\). Therefore, the vertical asymptote of this function is located at \(x = -\frac{1}{3}\). This means as \(x\) approaches \(-\frac{1}{3}\) from the right, the function value \(f(x)\) goes towards negative infinity, and we never actually cross this line on the graph.

• Vertical asymptote is defined by setting the inside of the logarithm to zero.
• For \(f(x) = \log(3x + 1)\), solve \(3x + 1 = 0\) to get the asymptote.
• Vertical asymptotes represent limitations in the domain of functions.

Logarithm Domain

The domain of a function is the set of all possible input values (\(x\)) that will produce valid output values (\(f(x)\)). For a logarithmic function like \(f(x)=\log(3x+1)\), we need to ensure that the expression inside the logarithm is positive.
The argument \(3x + 1\) must be greater than zero because the logarithm function is only defined for positive arguments. To find the domain:

• Solve the inequality \(3x + 1 > 0\).
• Subtract 1: \(3x > -1\).
• Divide by 3: \(x > -\frac{1}{3}\).

So, the domain of this function is all \(x\) values greater than \(-\frac{1}{3}\). In interval notation, this is expressed as \((-\frac{1}{3}, \infty)\). This tells us that any \(x\) less than \(-\frac{1}{3}\) is not included in the domain because it would make the argument non-positive, leading to an undefined logarithmic expression.

Solving Inequalities

Solving inequalities is a key skill in determining the domain of logarithmic functions and finding vertical asymptotes. Inequalities help define where certain conditions about a function's argument or output hold true. Let's break down the process using an example from our exercise:
To solve for the domain of \(f(x) = \log(3x + 1)\), we solve \(3x + 1 > 0\) which ensures the inside of the logarithm is positive.
Create a step-by-step solution:

• First, isolate the term with \(x\) on one side: \(3x + 1 > 0\).
• Subtract 1 from both sides to get \(3x > -1\).
• Finally, divide by 3: \(x > -\frac{1}{3}\).

Through these steps, we've shown how inequalities help to determine the set of inputs that a function can accept. In essence, not only do inequalities help in domain determination but also in identifying critical points where asymptotes exist.