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Chapter 3: Problem 61

Find the domain of each function. \(p(x)=\frac{x}{|2 x+3|-1}\)

Short Answer

Expert verified
The domain is \( \{ x \in \mathbb{R} \,|\, x e -1 \,\text{and}\, x e -2 \} \).

Step by step solution

01

- Identify the function form

Observe that the function is given as \( p(x) = \frac{x}{|2x+3| - 1} \). Recognize that the function consists of a numerator, \(x\), and a denominator, \(|2x + 3| - 1\).
02

- Determine when the denominator is defined

In order for the function \( p(x) \) to be defined, the denominator, \(|2x + 3| - 1\), must not be zero. Therefore, find values of \(x\) for which \(|2x + 3| - 1 = 0\).
03

- Solve the equation for the denominator

Set the denominator equal to zero and solve for \(x\):\[ |2x + 3| - 1 = 0 \]Adding 1 to both sides:\[ |2x + 3| = 1 \]This gives us two equations to solve:\[ 2x + 3 = 1 \] and \[ 2x + 3 = -1 \]
04

- Solve each equation separately

Solve the first equation:\[ 2x + 3 = 1 \]Subtract 3 from both sides:\[ 2x = -2 \]Divide by 2:\[ x = -1 \]Solve the second equation:\[ 2x + 3 = -1 \]Subtract 3 from both sides:\[ 2x = -4 \]Divide by 2:\[ x = -2 \]
05

- Combine the results

The values \( x = -1 \) and \( x = -2 \) make the denominator zero, causing the function to be undefined. Therefore, the domain of the function is all real numbers except \( x = -1 \) and \( x = -2 \).
06

- Write the domain in set notation

Express the domain of the function in set notation:\( \text{Domain}: \{ x \in \mathbb{R} \,|\, x e -1 \,\text{and}\, x e -2 \} \)

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Key Concepts

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

Denominator
In a fraction, the denominator is the bottom part of the fraction. It tells us how many parts the whole is divided into. For example, in the fraction \( \frac{3}{5} \), the denominator is \(5\). In this exercise, the function is \( p(x) = \frac{x}{|2x+3|-1} \). The denominator is \( |2x+3| - 1 \). The denominator cannot be zero because division by zero is undefined.
Absolute Value
Absolute value measures the distance a number is from zero on a number line, without considering direction. It's always a non-negative value. For example, \( | -4 | = 4 \) and \( | 4 | = 4 \), because both -4 and 4 are 4 units away from zero. In the function \( p(x) = \frac{x}{|2x+3|-1} \), \( |2x+3| \) means that whatever inside the absolute value bars, we take its non-negative distance from zero.
Set Notation
Set notation is a way to describe a set of numbers. When we talk about the domain of a function, we can use set notation to clearly list all possible values of \( x \). For example, if \( x \) can be any number except -1 and -2, we write this as: \( \{ x \in \mathbb{R} \,|\, x eq -1 \text{ and } x eq -2 \} \). This means 'the set of all real numbers \( x \) such that \( x \) is not equal to -1 and -2'.
Undefined Values
Values that make a function undefined are values for which the function cannot produce a result. For the function \( p(x) = \frac{x}{|2x+3|-1} \), if the denominator \( |2x + 3| - 1 \) equals zero, the function is undefined. Solving equations \( 2x + 3 = 1 \) and \( 2x + 3 = -1 \), we find \( x = -1 \) and \( x = -2 \) respectively. These values make the denominator zero, causing the function to be undefined. Hence, the domain excludes these values: the function is defined for all \( x \) except \( x = -1 \) and \( x = -2 \).

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Most popular questions from this chapter

The equation \(y=(x-c)^{2}\) defines a family of parabolas, one parabola for each value of \(c .\) On one set of coordinate axes, graph the members of the family for \(c=0, c=3\), and \(c=-2\)

Use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places. \(f(x)=-0.4 x^{4}-0.5 x^{3}+0.8 x^{2}-2 \quad[-3,2]\)

\(g(x)=x^{2}+1\) (a) Find the average rate of change from -1 to 2 . (b) Find an equation of the secant line containing \((-1, g(-1))\) and \((2, g(2))\).

The relationship between the Celsius \(\left({ }^{\circ} \mathrm{C}\right)\) and Fahrenheit \(\left({ }^{\circ} \mathrm{F}\right)\) scales for measuring temperature is given by the equation $$ F=\frac{9}{5} C+32 $$ The relationship between the Celsius \(\left({ }^{\circ} \mathrm{C}\right)\) and \(\mathrm{Kelvin}(\mathrm{K})\) scales is \(K=C+273 .\) Graph the equation \(F=\frac{9}{5} C+32\) using degrees Fahrenheit on the \(y\) -axis and degrees Celsius on the \(x\) -axis. Use the techniques introduced in this section to obtain the graph showing the relationship between Kelvin and Fahrenheit temperatures.

Solve: \(|3 x+7|-3=5\)
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