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Chapter 5: Problem 78

Make up an inequality that has no solution. Make up one that has exactly one solution.

Short Answer

Expert verified
Examples: 1. \[ x + 3 < x + 1 \]鈥� No solution.2. \[ 2(x - 1) = x + 1 \]鈥� Exactly one solution: \[ x = 3 \].

Step by step solution

01

Inequality with No Solution

One way to create an inequality with no solution is to present conflicting conditions. For example, consider the inequality \[ x + 3 < x + 1 \].We'll simplify this:
02

Simplifying the Inequality

Subtract x from both sides: \[ x + 3 - x < x + 1 - x \]This reduces to: \[ 3 < 1 \]Since 3 is never less than 1, this inequality has no solution.
03

Inequality with Exactly One Solution

Create an inequality where solving it results in a single value for x. For example, consider: \[ 2(x - 1) = x + 1 \].
04

Solving the Inequality

First, expand and move all terms involving x to one side:\[ 2x - 2 = x + 1 \]Subtract x from both sides:\[ x - 2 = 1 \]Add 2 to both sides:\[ x = 3 \]Thus, the inequality has exactly one solution: \[ x = 3 \].

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Key Concepts

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

inequalities with no solution
Inequalities can sometimes have no solutions. This happens when the conditions set by the inequality cannot ever be satisfied. For example, consider the inequality \[x + 3 < x + 1\].
This statement implies that 3 is less than 1, which is never true. When simplifying, you subtract x from both sides, resulting in \[3 < 1\].

Because this inequality will never hold true, there is no value for x that can satisfy it. Hence, it is an example of an inequality with no solution.
Understanding such cases helps avoid futile attempts at finding solutions.
solving linear inequalities
Solving linear inequalities involves isolating the variable on one side of the inequality sign. Let鈥檚 consider the inequality \[2(x - 1) = x + 1\].

First, expand the left-hand side: \[2x - 2\]. Next, subtract x from both sides: \[2x - 2 - x = x + 1 - x\], giving \[x - 2 = 1\].
Finally, add 2 to both sides to solve for x: \[x = 3\].

This method involves simple arithmetic steps:
  • Expand expressions.
  • Move terms involving x to one side.
  • Isolate x to find its value.
By following these steps, you can solve most linear inequalities efficiently.
single-variable inequalities
Single-variable inequalities involve expressions with only one variable. These are simpler to solve and understand. For example, if the inequality is \[x + 3 < x + 1\], you can subtract x from both sides:
\[ x + 3 - x < x + 1 - x \], yielding \[3 < 1\].

Since 3 is never less than 1, this inequality has no solution. Single-variable inequalities help build a foundation for understanding more complex multi-variable cases.
Common techniques in solving such inequalities include:
  • Combining like terms.
  • Isolating the variable.
  • Applying inverse operations (addition, subtraction, multiplication, or division).

Practice consistently to get comfortable with these basic steps.

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Most popular questions from this chapter

Originating on Pentecost Island in the Pacific, the practice of a person jumping from a high place harnessed to a flexible attachment was introduced to Western culture in 1979 by the Oxford University Dangerous Sport Club. One important parameter to know before attempting a bungee jump is the amount the cord will stretch at the bottom of the fall. The stiffness of the cord is related to the amount of stretch by the equation $$K=\frac{2 W(S+L)}{S^{2}}$$ where \(W=\) weight of the jumper (pounds) \(\begin{aligned} K &=\text { cord's stiffness (pounds per foot) } \\ L &=\text { free length of the cord (feet) } \\\ S &=\text { stretch (feet) } \end{aligned}\)

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find the function whose graph is the same as the graph of \(y=|x|\) but shifted down 4 units.

List the potential rational zeros of each polynomial function. Do not attempt to find the zeros. $$ f(x)=6 x^{4}+2 x^{3}-x^{2}+20 $$

We begin with two consecutive integers, \(a\) and \(a+1,\) for which \(f(a)\) and \(f(a+1)\) are of opposite sign. Evaluate \(f\) at the midpoint \(m_{1}\) of \(a\) and \(a+1 .\) If \(f\left(m_{1}\right)=0,\) then \(m_{1}\) is the zero of \(f,\) and we are finished. Otherwise, \(f\left(m_{1}\right)\) is of opposite sign to either \(f(a)\) or \(f(a+1) .\) Suppose that it is \(f(a)\) and \(f\left(m_{1}\right)\) that are of opposite sign. Now evaluate \(f\) at the midpoint \(m_{2}\) of \(a\) and \(m_{1} .\) Repeat this process until the desired degree of accuracy is obtained. Note that each iteration places the zero in an interval whose length is half that of the previous interval. Use the bisection method to approximate the zero of \(f(x)=8 x^{4}-2 x^{2}+5 x-1\) in the interval [0,1] correct to three decimal places. [Hint: The process ends when both endpoints agree to the desired number of decimal places.

Find bounds on the real zeros of each polynomial function. $$ f(x)=x^{4}+x^{3}-x-1 $$
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