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Find the domain of the function. $$ g(x)=\frac{2 x+1}{x-1} $$

Short Answer

Expert verified
The domain of the function \(g(x) = \frac{2x + 1}{x-1}\) is \(x \in (-\infty, 1) \cup (1, \infty)\), which includes all real numbers except \(x = 1\).

Step by step solution

01

Identify the problematic x-value(s)

First, we will identify the x-value(s) that make the denominator in the function equal to zero. For this, we will consider the expression in the denominator: $$ x - 1 = 0 $$
02

Solve for x

Next, we will solve the equation in step 1 to find the problematic x-value(s): $$ x = 1 $$ So, the function g(x) is undefined at x = 1 because it causes the denominator to become zero.
03

Determine the domain of the function

Now, based on the problematic x-value(s) found in step 2, we will determine the domain of the function. The domain of g(x) consists of all x-values except x = 1. So, we can represent the domain using interval notation as follows: $$ (-\infty, 1) \cup (1, \infty) $$ This means the domain of the function g(x) = (2x + 1)/(x-1) is all real numbers except x = 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Functions
A rational function is a type of function that can be expressed as the ratio of two polynomials. In simpler terms, it's a fraction where both the numerator and the denominator are polynomials. The general form of a rational function is \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomial expressions and \( Q(x) eq 0 \). Rational functions are significant in mathematics because they can model many real-world situations.

For example, the function \( g(x) = \frac{2x + 1}{x - 1} \) is a rational function because the numerator \( 2x + 1 \) and the denominator \( x - 1 \) are both polynomials. An important aspect of rational functions is that they are undefined when their denominator is zero. This is because division by zero is undefined in mathematics, which leads us to consider the notion of 'problematic x-values' in the next section.
Problematic X-Values
When analyzing rational functions, it is crucial to identify the problematic x-values. These are specific values for \( x \) that create a zero in the denominator of the function, leading to an undefined expression. In the expression \( g(x) = \frac{2x + 1}{x - 1} \), the denominator is \( x - 1 \), and setting it equal to zero, \( x - 1 = 0 \), reveals the problematic x-value to be \( x = 1 \).

Finding Problematic X-Values

To find these values, you simply solve the equation where the denominator equals zero. In our example, the following steps are undertaken:
  • Set the denominator equal to zero: \( x - 1 = 0 \)
  • Solve for \( x \) to find the problematic x-value: \( x = 1 \)
It's essential to exclude these values from the domain since including them would lead to division by zero, which is not allowable.
Interval Notation
To clearly express the set of values that are allowed for a function, interval notation is used. This notation is a way of writing subsets of the real number line. It communicates the starting and ending points of continuous intervals where a function is defined. For rational functions, interval notation allows us to define the domain, excluding the problematic x-values.

Let's look at the interval notation for the function \( g(x) = \frac{2x + 1}{x - 1} \). We know that \( x = 1 \) must be excluded. Using interval notation, we represent the domain of \( g(x) \) as \( (-\infty, 1) \cup (1, \infty) \), which reads as 'all real numbers except x = 1'. Here's a quick breakdown:
  • The parenthesis \( ( \) or \( ) \) indicate that the endpoint is not included in the interval, also known as an open interval.
  • The union symbol \( \cup \) indicates that all the numbers in both intervals are included in the domain, except for the excluded problematic x-value \( x = 1 \).
Interval notation is a succinct and precise way to describe the domain of functions, and it is especially handy when dealing with rational functions and their restrictions.

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