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Chapter 7: Problem 12

For each rational function, find all numbers that are not in the domain. Then give the domain, using set-builder notation. $$ f(x)=\frac{9 x^{2}-8 x+3}{4 x^{2}+1} $$

Short Answer

Expert verified
The domain is all real numbers: \(\{x \mid x \in \mathbb{R}\}\). There are no numbers that are not in the domain.

Step by step solution

01

Identify the Denominator

Analyze the rational function and focus on the denominator. The denominator of the given function is \(4x^2 + 1\).
02

Set the Denominator Equal to Zero

Set the denominator equal to zero to find the values of \(x\) that make the denominator zero. Solve the equation \(4x^2 + 1 = 0\).
03

Solve for x

Solve the equation \(4x^2 + 1 = 0\). Subtract 1 from both sides to get \(4x^2 = -1\). Now, divide both sides by 4 to get \(x^2 = -\frac{1}{4}\).
04

Analyze the Solution

Since \(x^2 = -\frac{1}{4}\) does not have any real solutions (as a square of a real number cannot be negative), there are no real roots. Hence, the denominator never equals zero for real numbers.
05

Establish the Domain

Since there are no values of \(x\) that make the denominator zero, the function is defined for all real numbers. Thus, the domain of the function is all real numbers.

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Key Concepts

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

Denominator Analysis
Let's begin by focusing on the denominator analysis. We need to identify which part of the rational function could potentially cause problems for us by making the fraction undefined. In the given problem, the function is:

\[ f(x)=\frac{9x^{2}-8x+3}{4x^{2}+1} \]
The denominator here is \[ 4x^{2}+1 \]. For a rational function to be undefined, the denominator must equal zero. Our goal is to find out if there are any values of \( x \) that can make the denominator equal zero.
Solving Equations
To discover the values of \( x \) that make the denominator equal to zero, let's set the denominator equal to zero and solve the equation: \[ 4x^{2} + 1 = 0 \]
Now, solve for \( x \). Start by subtracting 1 from both sides:
\[ 4x^{2} = -1 \]
Next, divide both sides by 4 to isolate \( x^2 \):
\[ x^{2} = -\frac{1}{4} \]
Notice that the equation \( x^{2} = -\frac{1}{4} \) has no real solutions, because you cannot take the square root of a negative number within the real number system. This tells us that there are no real values of \( x \) that make the denominator zero.
Set-Builder Notation
Now that we know the denominator does not become zero for any real numbers, we can conclude that the domain of the function is all real numbers. To express this using set-builder notation, we write:

\[ \text{Domain} = \{ x \in \mathbb{R} \} \]
This notation reads 'the set of all \( x \) such that \( x \) is a real number.' Thus, the function \( f(x) \) is defined for all real numbers.

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