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Chapter 0: Problem 44

Find the domain of each function given below. $$f(x)=\frac{2 x-1}{9-2 x}$$

Short Answer

Expert verified
The domain is \( (-\infty, \frac{9}{2}) \cup (\frac{9}{2}, \infty) \).

Step by step solution

01

- Identify the function type

Determine if the given function is a rational function. The given function is \( f(x) = \frac{2x - 1}{9 - 2x} \), which is a rational function because it is in the form \( \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials.
02

- Set the denominator equal to zero

For the domain of a rational function, identify values of \( x \) that make the denominator zero, as the function is undefined at these points. Set the denominator \( 9 - 2x \) to zero and solve for \( x \):
03

- Solve the equation

Solve the equation \( 9 - 2x = 0 \): \[ 9 - 2x = 0 \] \[ -2x = -9 \] \[ x = \frac{9}{2} \]. So, \( x = \frac{9}{2} \) makes the denominator zero.
04

- Determine the domain

Exclude the value \( x = \frac{9}{2} \) from the domain of the function. The domain of \( f(x) = \frac{2x - 1}{9 - 2x} \) consists of all real numbers except \( x = \frac{9}{2} \). In interval notation, the domain is: \( (-\infty, \frac{9}{2}) \cup (\frac{9}{2}, \infty) \)

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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 expressed as a fraction where both the numerator and the denominator are polynomials. For example, in the function given in the exercise, \(f(x) = \frac{2x - 1}{9 - 2x}\), the numerator is \(2x - 1\) and the denominator is \(9 - 2x\).
Because rational functions are ratios of polynomials, they are treated differently compared to regular polynomials.
One critical aspect of rational functions is determining their domains, which leads us to the next important concept.
Domain Exclusion
Domain exclusion refers to finding values that make the function undefined. For rational functions, the function is undefined wherever the denominator is zero. In simpler terms, any value of \(x\) that makes the denominator zero must be excluded from the domain.
In the exercise, we identified that setting the denominator \(9 - 2x\) to zero and solving gives \(x = \frac{9}{2}\).
This means the function \(f(x) = \frac{2x - 1}{9 - 2x}\) is undefined when \(x = \frac{9}{2}\).
These values are excluded to find the valid, acceptable range of the function, leading us to the interval notation.
Interval Notation
Interval notation is a way to signify the set of all acceptable input values (domain) for a function. In the notation, parentheses \( ( ) \) denote that an endpoint is not included, and brackets \( [ ] \) denote inclusion.
In our function, \(f(x) = \frac{2x - 1}{9 - 2x}\), we need to exclude \(x = \frac{9}{2}\) from the domain. To express all other numbers except for \(x = \frac{9}{2}\), we use interval notation as follows:
\( (-\infty, \frac{9}{2}) \cup (\frac{9}{2}, \infty) \).

This means that the domain includes all real numbers less than \(\frac{9}{2}\) and all real numbers greater than \(\frac{9}{2}\), but \(\frac{9}{2}\) itself is not included.

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Most popular questions from this chapter

The stopping distance (at some fixed speed) of regular tires on glare ice is given by a linear function of the air temperature \(F\) \(D(F)=2 F+115\) where \(D(F)\) is the stopping distance, in feet, when the air temperature is \(F\), in degrees Fahrenheit. a) Find \(D\left(0^{\circ}\right), D\left(-20^{\circ}\right), D\left(10^{\circ}\right),\) and \(D\left(32^{\circ}\right).\) b) Explain why the domain should be restricted to the interval \(\left[-57.5^{\circ}, 32^{\circ}\right].\)

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