Problem 3
Question
Solve each inequality and graph its solution set on a number line. $$(x+1)(x+4)<0$$
Step-by-Step Solution
Verified Answer
The solution is \((-4, -1)\), and it is graphed as an open interval on the number line.
1Step 1: Identify Critical Points
Set each factor of the inequality \((x+1)(x+4) < 0\) to zero to find the critical points. \(x+1 = 0\) gives \(x = -1\) and \(x+4 = 0\) gives \(x = -4\). The critical points divide the number line into intervals.
2Step 2: Test Intervals
The critical points \(-4\) and \(-1\) divide the number line into three intervals: \((-\infty, -4)\), \((-4, -1)\), and \((-1, \infty)\). Choose a test point from each interval to determine where the inequality is satisfied. - For \(x = -5\) in \((-\infty, -4)\): \((-5+1)(-5+4) = (-4)(-1) = 4\). This is positive, so the inequality is not satisfied.- For \(x = -2\) in \((-4, -1)\): \((-2+1)(-2+4) = (-1)(2) = -2\). This is negative, so the inequality is satisfied.- For \(x = 0\) in \((-1, \infty)\): \((0+1)(0+4) = (1)(4) = 4\). This is positive, so the inequality is not satisfied.
3Step 3: Write the Solution
Since the inequality \((x+1)(x+4) < 0\) is satisfied only within the interval \((-4, -1)\), this is the solution set. The solution is an open interval because neither \(-1\) nor \(-4\) makes the inequality true.
4Step 4: Graph the Solution
Draw a number line and place open circles at \(-4\) and \(-1\) since these points are not included in the solution. Shade the region between \(-4\) and \(-1\) to represent the interval \((-4, -1)\) where the inequality holds true.
Key Concepts
Critical Points in InequalitiesInterval TestingGraphing Solution SetsOpen IntervalsNumber Line Representation
Critical Points in Inequalities
When solving inequalities, critical points serve as boundaries that can change the sign of the expression. For an inequality like \[(x+1)(x+4) < 0\]we begin by finding these points. Set the expression equal to zero:
- \(x+1=0\) results in \(x=-1\)
- \(x+4=0\) results in \(x=-4\)
Interval Testing
Interval testing is the method used to check which sections defined by critical points satisfy the inequality. After identifying critical points, the number line is split into intervals:
- \((-\infty, -4)\)
- \((-4, -1)\)
- \((-1, \infty)\)
- For \(x = -5\) in \((-\infty, -4)\), the expression is positive.
- For \(x = -2\) in \((-4, -1)\), it is negative. Thus, the inequality is satisfied here.
- For \(x = 0\) in \((-1, \infty)\), the result turns positive again.
Graphing Solution Sets
Visualizing solution sets of inequalities on a graph, specifically a number line, enhances comprehension. Once the valid interval is determined, mark it:
- Use open circles where critical points occur, \(-4\) and \(-1\).
- Draw a line to connect these points.
- Shade the line between \(-4\) and \(-1\) to indicate where the inequality holds true.
Open Intervals
Solution sets of certain inequalities are termed open intervals. An open interval does not include its endpoints. For this inequality, \((x+1)(x+4) < 0\)tests reveal the inequality holds between critical points \(-4\) and \(-1\), but not at them. Hence, the solution is \((-4, -1)\).
- The endpoints \(-4\) and \(-1\) are not part of the solution, indicated by the lack of equality in the inequality symbol "<".
- Such cases require using open circles in the number line graph.
Number Line Representation
Using a number line is an effective way to represent solutions visually. To construct one for this inequality:
- Draw a horizontal line split into equal sections.
- Mark critical points with open circles to signify they aren’t included in the solution.
- Shade the interval between \(-4\) and \(-1\) to highlight where the inequality \((x+1)(x+4) < 0\)is true.
Other exercises in this chapter
Problem 2
Solve each of the quadratic equations by factoring and applying the property, \(a b=0\) if and only if \(a=0\) or \(b=0\). If necessary, return to Chapter 3 and
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Label each statement true or false. Every real number is a complex number.
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Solve each quadratic equation using the method that seems most appropriate to you. $$3 x^{2}+23 x-36=0$$
View solution Problem 3
First use the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real soluti
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