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Chapter 1: Problem 52

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the interval(s) on a number line. $$|x|>0$$

Short Answer

Expert verified
The set of real numbers satisfying \(|x| > 0\) is \((-\infty, 0) \cup (0, \infty)\) and excludes zero.

Step by step solution

01

Understand Absolute Value

The given inequality is \(|x| > 0\). The absolute value of a number, \(|x|\), is its distance from 0 on the number line, without considering direction. This means \(|x| > 0\) if and only if \(x\) is not equal to 0, because the distance from zero for any non-zero number is positive.
02

Solve the Inequality

The inequality \(|x| > 0\) implies that \(x\) must be either positive or negative, but not zero. This translates to the set of solutions \(x \in (-\infty, 0) \cup (0, \infty)\). Essentially, \(x\) can be any real number except zero.
03

Express the Solution in Interval Notation

In interval notation, \(x \in (-\infty, 0) \cup (0, \infty)\) suggests two open intervals because 0 is not included in the set. We use parentheses to denote open intervals, which indicate that the endpoints are not part of the solution.
04

Graph the Solution on a Number Line

To represent the solution graphically, draw a number line and shade all the points along the line except for the point at zero. Place open circles at zero to show that it is not included.

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Key Concepts

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

Real Numbers
Real numbers include all the possible numbers you can think of. They cover both rational numbers, like fractions and whole numbers, and irrational numbers, like square roots that can't be simplified. You can picture real numbers as a continuous line stretching in both directions without ending.Real numbers are vital in understanding absolute value inequalities. An inequality such as \( |x| > 0 \) means you are looking for all real numbers except zero. In this situation, none of the rational or irrational numbers, apart from zero, fulfill the inequality. This is because zero is the only real number whose absolute value is not greater than itself.When solving absolute value inequalities, it's important to remember that this set includes all these types of numbers as potential solutions. Knowing this helps when identifying which values might satisfy the given inequality.
Interval Notation
Interval notation is a way to express a range of numbers along a number line. It gives you a neat way to write where a solution begins and ends. For the inequality \( |x| > 0 \), we are interested in values on either side of zero, which the inequality does not include.In interval notation, we express this solution as \((-\infty, 0) \cup (0, \infty)\). Here's how this works:
  • \((-\infty, 0)\): This notation covers all numbers less than zero. The parentheses indicate that zero is not included.
  • \((0, \infty)\): This covers all numbers greater than zero, again excluding zero itself.
  • The operator \(\cup\) signifies the union of these two sets, meaning any number in either interval is a solution.
This notation is concise and shows at a glance that every number except zero satisfies the inequality.
Number Line Graphing
Graphing on a number line is a useful visual way to demonstrate which numbers are solutions to an inequality. For the inequality \( |x| > 0 \), graphing involves marking all real numbers except zero.Here's how to graph this:
  • Draw a horizontal line to represent your number line.
  • Mark a circle on zero. This indicates zero is not part of the solution.
  • Shade the entire line except for zero. This shows all negative and positive numbers are included in the solution.
  • The circles on zero remain open, highlighting the absence of this number in the solution set.
A number line is a very intuitive way to visualize solutions, especially when you're dealing with inequalities where certain values are excluded, like in this case with zero.

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Most popular questions from this chapter

In each of parts (a) through (d), first solve the equation for \(y\) so that you can enter it in your graphing utility. Then use the graphing utility to graph the equation in an appropriate viewing rectangle. In each case, the graph is a line. Given that the \(x\) - and \(y\) -intercepts are (in every case here) integers, read their values off the screen and write them down for easy reference when you get to part (e). (a) \(\frac{x}{2}+\frac{y}{3}=1\) (c) \(\frac{x}{6}+\frac{y}{5}=1\) (b) \(\frac{x}{-2}+\frac{y}{-3}=1\) (d) \(\frac{x}{-6}+\frac{y}{-5}=1\) (e) On the basis of your results in parts (a) through (d), describe, in general, the graph of the equation \(\frac{x}{a}+\frac{y}{b}=1,\) where \(a\) and \(b\) are nonzero constants.

Use a graphing utility to graph the equations and to approximate the \(x\) -intercepts. In approximating the \(x\) -intercepts, use a "solve" key or a sufficiently magnified view to ensure that the values you give are correct in the first three decimal places. Remark: None of the \(x\) -intercepts for these four equations can be obtained using factoring techniques.) $$y=8 x^{3}-6 x-1$$

Imagine that you own a grove of orange trees, and suppose that from past experience you know that when 100 trees are planted, each tree will yield approximately 240 oranges per year. Furthermore, you've noticed that when additional trees are planted in the grove, the yield per tree decreases. Specifically, you have noted that the yield per tree decreases by about 20 oranges for each additional tree planted. (a) Let \(y\) denote the yield per tree when \(x\) trees are planted. Find a linear equation relating \(x\) and \(y\) Hint: You are given that the point (100,240) is on the line. What is given about \(\Delta y / \Delta x ?\) (b) Use the equation in part (a) to determine how many trees should be planted to obtain a yield of 400 oranges per tree. (c) If the grove contains 95 trees, what yield can you expect from each tree?

This exercise outlines a proof of the fact that two nonvertical lines with slopes \(m_{1}\) and \(m_{2}\) are perpendicular if and only if \(m_{1} m_{2}=-1 .\) In the following figure, we've assumed that our two nonvertical lines \(y=m_{1} x\) and \(y=m_{2} x\) intersect at the origin. [If they did not intersect there, we could just as well work with lines parallel to these that do intersect at \((0,0),\) recalling that parallel lines have the same slope.] The proof relies on the following geometric fact: \(\overline{O A} \perp \overline{O B} \quad\) if and only if \(\quad(O A)^{2}+(O B)^{2}=(A B)^{2}\). (a) Verify that the coordinates of \(A\) and \(B\) are \(A\left(1, m_{1}\right)\) and \(B\left(1, m_{2}\right)\) (b) Show that $$\begin{aligned}O A^{2} &=1+m_{1}^{2} \\\O B^{2} &=1+m_{2}^{2} \\\A B^{2} &=m_{1}^{2}-2 m_{1} m_{2}+m_{2}^{2} \end{aligned}$$ (c) Use part (b) to show that the equation $$O A^{2}+O B^{2}=A B^{2}$$ is equivalent to \(m_{1} m_{2}=-1\) (GRAPH CAN'T COPY)

Rewrite each statement using absolute value notation, as in Example 5. The number \(y\) is less than one unit from the number \(t\).
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