2.4: Inequalities with Absolute Value and Quadratic Functions

Math 370 Learning Objectives

The topics in this section are left for the student to review, as needed; however, the final topic (Regions in Planes) might be covered in class despite not being a topic within the Math 370 curriculum. The reason for this is that this skill is needed in Calculus II.

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Solving Linear Equations and Inequalities Graphically

In this section, not only do we develop techniques for solving various classes of inequalities analytically,¹ we also look at them graphically. The first example motivates the core ideas.

Example \( \PageIndex{1} \)

Let \(f(x) = 2x-1\) and \(g(x)=5\).

1. Solve \(f(x) = g(x)\).
2. Solve \(f(x) < g(x)\).
3. Solve \(f(x) > g(x)\).
4. Graph \(y=f(x)\) and \(y=g(x)\) on the same set of axes and interpret your solutions to parts a through c above.

Solution

1. To solve \(f(x) = g(x)\), we replace \(f(x)\) with \(2x-1\) and \(g(x)\) with \(5\) to get \(2x-1 = 5\). Solving for \(x\), we get \(x=3\).

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2. The inequality \(f(x) < g(x)\) is equivalent to \(2x-1 < 5\). Solving gives \(x < 3\) or \((-\infty, 3)\).

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3. To find where \(f(x) > g(x)\), we solve \(2x-1 > 5\). We get \(x > 3\), or \((3, \infty)\).

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4. To graph \(y=f(x)\), we graph \(y = 2x-1\), which is a line with a \(y\)-intercept of \((0,-1)\) and a slope of \(2\). The graph of \(y=g(x)\) is \(y=5\) which is a horizontal line through \((0,5)\) (see Figure \( \PageIndex{1} \).

   
   Figure \( \PageIndex{1} \)

   To see the connection between the graph and the Algebra, we recall the Fundamental Graphing Principle for Functions in Section 1.6: the point \((a,b)\) is on the graph of \(f\) if and only if \(f(a)=b\). In other words, a generic point on the graph of \(y=f(x)\) is \((x,f(x))\), and a generic point on the graph of \(y=g(x)\) is \((x,g(x))\). When we seek solutions to \(f(x)=g(x)\), we are looking for \(x\) values whose \(y\) values on the graphs of \(f\) and \(g\) are the same.
   
   In part a, we found \(x=3\) is the solution to \(f(x)=g(x)\). Sure enough, \(f(3) = 5\) and \(g(3) = 5\) so that the point \((3,5)\) is on both graphs. In other words, the graphs of \(f\) and \(g\) intersect at \((3,5)\).
   
   In part b, we set \(f(x) < g(x)\) and solved to find \(x < 3\). For \(x < 3\), the point \((x,f(x))\) is below \((x,g(x))\) since the \(y\) values on the graph of \(f\) are less than the \(y\) values on the graph of \(g\) there.
   
   Analogously, in part c, we solved \(f(x) > g(x)\) and found \(x > 3\). For \(x > 3\), note that the graph of \(f\) is above the graph of \(g\), since the \(y\) values on the graph of \(f\) are greater than the \(y\) values on the graph of \(g\) for those values of \(x\) (see Figure \( \PageIndex{2} \).

   
   Figure \( \PageIndex{2} \)

The preceding example demonstrates the following, which is a consequence of the Fundamental Graphing Principle for Functions.

Theorem \(\PageIndex{1}\): Graphical Interpretation of Solutions to Equations and Inequalities

Suppose \(f\) and \(g\) are functions.

• The solutions to \(f(x)=g(x)\) are the \(x\) values where the graphs of \(y=f(x)\) and \(y=g(x)\) intersect.
• The solution to \(f(x) < g(x)\) is the set of \(x\) values where the graph of \(y=f(x)\) is below the graph of \(y=g(x)\).
• The solution to \(f(x) > g(x)\) is the set of \(x\) values where the graph of \(y=f(x)\) above the graph of \(y=g(x)\).

Subsection Footnotes

¹ Remember, "analytically" means "by hand."

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Finding Solutions to Equations and Inequalities Given Graphs

The next example turns the tables and furnishes the graphs of two functions and asks for solutions to equations and inequalities.

Example \( \PageIndex{2} \)

The graphs of \(f\) and \(g\) are in Figure \( \PageIndex{3} \)below. (The graph of \(y=g(x)\) is bolded.) Use these graphs to answer the following questions.

Figure \( \PageIndex{3} \)

1. Solve \(f(x) = g(x)\).
2. Solve \(f(x) < g(x)\).
3. Solve \(f(x) \geq g(x)\).

Solution

1. To solve \(f(x)=g(x)\), we look for where the graphs of \(f\) and \(g\) intersect. These appear² to be at the points \((-1,2)\) and \((1,2)\), so our solutions to \(f(x) = g(x)\) are \(x = -1\) and \(x=1\).

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2. To solve \(f(x) < g(x)\), we look for where the graph of \(f\) is below the graph of \(g\). This appears to happen for the \(x\) values less than \(-1\) and greater than \(1\). Our solution is \((-\infty, -1) \cup (1,\infty)\).

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3. To solve \(f(x) \geq g(x)\), we look for solutions to \(f(x)=g(x)\) as well as \(f(x) > g(x)\). We solved the former equation and found \(x = \pm 1\). To solve \(f(x) > g(x)\), we look for where the graph of \(f\) is above the graph of \(g\). This appears to happen between \(x=-1\) and \(x=1\), on the interval \((-1,1)\). Hence, our solution to \(f(x) \geq g(x)\) is \([-1,1]\) (see Figure \( \PageIndex{4} \)).

Figure \( \PageIndex{4} \)

Subsection Footnotes

² I use the word "appear" here because we cannot guarantee our solutions are correct by "eyeballing" them from the graph.

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Solving Absolute Value Inequalities

We now turn our attention to solving inequalities involving the absolute value. We have the following theorem from Intermediate Algebra to help us.

Theorem \(\PageIndex{2}\): Inequalities Involving the Absolute Value

Let \(c\) be a real number.

• For \(c > 0\), \(|x| < c\) is equivalent to \(-c<x<c\).
• For \(c > 0\), \(|x| \leq c\) is equivalent to \(-c \leq x \leq c\).
• For \(c \leq 0\), \(|x| < c\) has no solution, and for \(c < 0\), \(|x| \leq c\) has no solution.
• For \(c \geq 0\), \(|x| > c\) is equivalent to \(x<-c\) or \(x>c\).
• For \(c \geq 0\), \(|x| \geq c\) is equivalent to \(x \leq -c\) or \(x \geq c\).
• For \(c < 0\), \(|x| > c\) and \(|x| \geq c\) are true for all real numbers.

As with our proof of the Properties of the Absolute Value in Section 2.2, we could argue Theorem \( \PageIndex{2} \) using cases. However, in light of what we have developed in this section, we can understand these statements graphically. For instance, if \(c > 0\), the graph of \(y=c\) is a horizontal line which lies above the \(x\)-axis through \((0,c)\). To solve \(|x| < c\), we are looking for the \(x\) values where the graph of \(y=|x|\) is below the graph of \(y=c\). We know that the graphs intersect when \(|x|=c\), which, from Section 2.2, we know happens when \(x=c\) or \(x=-c\). Graphing, we get

Figure \( \PageIndex{5} \)

We see that the graph of \(y=|x|\) is below \(y=c\) for \(x\) between \(-c\) and \(c\), and hence we get \(|x| < c\) is equivalent to \(-c < x < c\). The other properties in Theorem \( \PageIndex{2} \) can be shown similarly.

Example \( \PageIndex{3} \)

Solve the following inequalities analytically; check your answers graphically.

1. \(|x-1| \geq 3\)
2. \(4 - 3|2x+1| > -2\)
3. \(2 < |x-1| \leq 5\)
4. \(|x+1|\geq \frac{x+4}{2}\)

Solution

1. From Theorem \( \PageIndex{2} \), \(|x-1|\geq3\) is equivalent to \(x-1 \leq -3\) or \(x-1 \geq 3\). Solving, we get \(x \leq -2\) or \(x \geq 4\), which, in interval notation is \((-\infty,-2] \cup [4,\infty)\). Graphically, we have

   
   Figure \( \PageIndex{6} \)

   We see that the graph of \(y=|x-1|\) is above the horizontal line \(y=3\) for \(x < -2\) and \(x > 4\) hence this is where \(|x-1| > 3\). The two graphs intersect when \(x=-2\) and \(x=4\), so we have graphical confirmation of our analytic solution.

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2. To solve \(4 - 3|2x+1| > -2\) analytically, we first isolate the absolute value before applying Theorem \( \PageIndex{2} \). To that end, we get \(-3|2x+1|>-6\) or \(|2x+1|<2\). Rewriting, we now have \(-2 < 2x+1 < 2\) so that \(-\frac{3}{2} < x < \frac{1}{2}\). In interval notation, we write \(\left(-\frac{3}{2}, \frac{1}{2}\right)\). Graphically we see that the graph of \(y=4-3|2x+1|\) is above \(y=-2\) for \(x\) values between \(-\frac{3}{2}\) and \(\frac{1}{2}\).

   
   Figure \( \PageIndex{7} \)

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3. Rewriting the compound inequality \(2 < |x-1| \leq 5\) as "\(2 < |x-1|\) and \(|x-1| \leq 5\)" allows us to solve each piece using Theorem \( \PageIndex{2} \). The first inequality, \(2 < |x-1|\) can be re-written as \(|x-1|>2\) so \(x-1 < -2\) or \(x-1 > 2\). We get \(x<-1\) or \(x>3\). Our solution to the first inequality is then \((-\infty, -1) \cup (3, \infty)\). For \(|x-1| \leq 5\), we get \(-5 \leq x-1 \leq 5\) so that \(-4 \leq x \leq 6\), or \([-4,6]\). Our solution to \(2 < |x-1| \leq 5\) is comprised of values of \(x\) which satisfy both parts of the inequality, so we take the intersection of \((-\infty, -1) \cup (3, \infty)\) and \([-4,6]\) to get \([-4,-1) \cup (3,6]\). Graphically, we see that the graph of \(y=|x-1|\) is "between" the horizontal lines \(y=2\) and \(y=5\) for \(x\) values between \(-4\) and \(-1\) as well as those between \(3\) and \(6\). Including the \(x\) values where \(y=|x-1|\) and \(y=5\) intersect, we get

   
   Figure \( \PageIndex{8} \)

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4. We need to exercise some special caution when solving \(|x+1|\geq \frac{x+4}{2}\). As we saw in Example \( \PageIndex{1} \) in Section 2.2, when variables are both inside and outside of the absolute value, it’s usually best to refer to the definition of absolute value to remove the absolute values and proceed from there. To that end, we have \(|x+1| = -(x+1)\) if \(x < -1\) and \(|x+1| = x+1\) if \(x \geq -1\).
   
   We break the inequality into cases, the first case being when \(x<-1\). For these values of \(x\), our inequality becomes \(-(x+1) \geq \frac{x+4}{2}\). Solving, we get \(-2x-2 \geq x+4\), so that \(-3x \geq 6\), which means \(x \leq -2\). Since all of these solutions fall into the category \(x < -1\), we keep them all.
   
   For the second case, we assume \(x \geq -1\). Our inequality becomes \(x+1 \geq \frac{x+4}{2}\), which gives \(2x+2 \geq x+4\) or \(x \geq 2\). Since all of these values of \(x\) are greater than or equal to \(-1\), we accept all of these solutions as well. Our final answer is \((-\infty, -2] \cup [2,\infty)\).

   
   Figure \( \PageIndex{9} \)

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Solving Quadratic Inequalities

We now turn our attention to quadratic inequalities. In the last example of Section 2.3, we needed to determine the solution to \(x^2 - x -6 < 0\). We will now revisit this problem using some of the techniques developed in this section not only to reinforce our solution in Section 2.3, but to also help formulate a general analytic procedure for solving all quadratic inequalities.

If we consider \(f(x) = x^2-x-6\) and \(g(x)=0\), then solving \(x^2 - x -6 < 0\) corresponds graphically to finding the values of \(x\) for which the graph of \(y=f(x)=x^2-x-6\) (the parabola) is below the graph of \(y=g(x)=0\) (the \(x\)-axis). We’ve provided the graph again for reference.

Figure \( \PageIndex{10} \)

We can see that the graph of \(f\) does dip below the \(x\)-axis between its two \(x\)-intercepts. The zeros of \(f\) are \(x=-2\) and \(x=3\) in this case and they divide the domain (the \(x\)-axis) into three intervals: \((-\infty, -2)\), \((-2,3)\) and \((3, \infty)\). For every number in \((-\infty, -2)\), the graph of \(f\) is above the \(x\)-axis; in other words, \(f(x) > 0\) for all \(x\) in \((-\infty, -2)\). Similarly, \(f(x) < 0\) for all \(x\) in \((-2,3)\), and \(f(x) > 0\) for all \(x\) in \((3, \infty)\). We can schematically represent this with a sign diagram (see Figure \( \PageIndex{11} \)).

Figure \( \PageIndex{11} \)

Here, the \((+)\) above a portion of the number line indicates \(f(x) > 0\) for those values of \(x\); the \((-)\) indicates \(f(x) < 0\) there. The numbers labeled on the number line are the zeros of \(f\), so we place \(0\) above them. We see at once that the solution to \(f(x) < 0\) is \((-2,3)\).

An important property of quadratic functions is that if the function is positive at one point and negative at another, the function must have at least one zero in between.³ Graphically, this means that a parabola can’t be above the \(x\)-axis at one point and below the \(x\)-axis at another point without crossing the \(x\)-axis. This allows us to determine the sign of all of the function values on a given interval by testing the function at just one value in the interval. We use this idea in the following example.

Example \( \PageIndex{4} \)

Solve the following inequalities analytically using sign diagrams. Verify your answer graphically.

1. \(2x^2 \leq 3-x\)
2. \(x^2 - 2x > 1\)
3. \(2x-x^2 \geq |x-1|-1\)

Solution

1. To solve \(2x^2 \leq 3-x\), we first get \(0\) on one side of the inequality which yields \(2x^2+x-3 \leq 0\). We find the zeros of \(f(x) = 2x^2 + x - 3\) by solving \(2x^2 + x - 3 = 0\) for \(x\). Factoring gives \((2x+3)(x-1)=0\), so \(x = -\frac{3}{2}\) or \(x = 1\). We place these values on the number line with \(0\) above them and choose test values in the intervals \(\left(-\infty, -\frac{3}{2}\right)\), \(\left(-\frac{3}{2},1\right)\) and \((1,\infty)\). For the interval \(\left(-\infty, -\frac{3}{2}\right)\), we choose \(x=-2\); for \(\left(-\frac{3}{2},1\right)\), we pick \(x=0\); and for \((1,\infty)\), \(x=2\). Evaluating the function at the three test values gives us \(f(-2) = 3 > 0\), so we place \((+)\) above \(\left(-\infty, -\frac{3}{2}\right)\); \(f(0)=-3 < 0\), so \((-)\) goes above the interval \(\left(-\frac{3}{2},1\right)\); and, \(f(2) = 7\), which means \((+)\) is placed above \((1,\infty)\). See Figure \( \PageIndex{12}A \).
   
   Since we are solving \(2x^2+x-3 \leq 0\), we look for solutions to \(2x^2+x-3 < 0\) as well as solutions for \(2x^2+x-3 =0\). For \(2x^2+x-3 < 0\), we need the intervals which we have a \((-)\). Checking the sign diagram, we see this is \(\left(-\frac{3}{2},1\right)\). We know \(2x^2+x-3 =0\) when \(x=-\frac{3}{2}\) and \(x=1\), so our final answer is \(\left[-\frac{3}{2},1\right]\).
   
   To verify our solution graphically, we refer to the original inequality, \(2x^2 \leq 3-x\). We let \(g(x) = 2x^2\) and \(h(x)=3-x\). We are looking for the \(x\) values where the graph of \(g\) is below that of \(h\) (the solution to \(g(x) < h(x)\)) as well as the points of intersection (the solutions to \(g(x)=h(x)\)). See Figure \( \PageIndex{12}B \).

   
   Figures \( \PageIndex{12}A \) (left) and \( \PageIndex{12}B \) (right)

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2. Once again, we re-write \(x^2-2x > 1\) as \(x^2-2x-1>0\) and we identify \(f(x)=x^2-2x-1\). When we go to find the zeros of \(f\), we find, to our chagrin, that the quadratic \(x^2-2x-1\) doesn’t factor nicely. Hence, we resort to the Quadratic Formula to solve \(x^2-2x-1=0\), and arrive at \(x=1 \pm \sqrt{2}\). As before, these zeros divide the number line into three pieces. To help us decide on test values, we approximate \(1 - \sqrt{2} \approx -0.4\) and \(1 + \sqrt{2} \approx 2.4\). We choose \(x=-1\), \(x=0\) and \(x=3\) as our test values and find \(f(-1)= 2\), which is \((+)\); \(f(0)=-1\) which is \((-)\); and \(f(3)=2\) which is \((+)\) again (see Figure \( \PageIndex{13}A \)). Our solution to \(x^2-2x-1>0\) is where we have \((+)\), so, in interval notation \(\left(-\infty, 1-\sqrt{2}\right) \cup \left(1+\sqrt{2},\infty\right)\).
   
   To check the inequality \(x^2 - 2x > 1\) graphically, we set \(g(x) = x^2-2x\) and \(h(x)=1\). We are looking for the \(x\) values where the graph of \(g\) is above the graph of \(h\) (see Figure \( \PageIndex{13}B \)).

   
   Figures \( \PageIndex{13}A \) (left) and \( \PageIndex{13}B \) (right)

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3. To solve our last inequality, \(2x-x^2 \geq |x-1|-1\), we re-write the absolute value using cases.
   
   For \(x < 1\), \(|x-1| = -(x-1) = 1-x\), so we get \(2x-x^2 \geq 1-x-1\), or \(x^2-3x \leq 0\). Finding the zeros of \(f(x) = x^2-3x\), we get \(x=0\) and \(x=3\). However, we are only concerned with the portion of the number line where \(x < 1\), so the only zero that we concern ourselves with is \(x=0\). This divides the interval \(x<1\) into two intervals: \((-\infty, 0)\) and \((0,1)\). We choose \(x=-1\) and \(x=\frac{1}{2}\) as our test values. We find \(f(-1) = 4\) and \(f\left(\frac{1}{2}\right) = -\frac{5}{4}\). Hence, our solution to \(x^2-3x \leq 0\) for \(x < 1\) is \([0,1)\).
   
   Next, we turn our attention to the case \(x \geq 1\). Here, \(|x-1| = x-1\), so our original inequality becomes \(2x-x^2 \geq x-1-1\), or \(x^2-x-2 \leq 0\). Setting \(g(x) = x^2-x-2\), we find the zeros of \(g\) to be \(x=-1\) and \(x=2\). Of these, only \(x=2\) lies in the region \(x \geq 1\), so we ignore \(x=-1\). Our test intervals are now \([1,2)\) and \((2,\infty)\). We choose \(x=1\) and \(x=3\) as our test values and find \(g(1) = -2\) and \(g(3) = 4\). Hence, our solution to \(g(x) = x^2-x-2 \leq 0\), in this region is \([1,2)\).Combining these into one sign diagram, we have that our solution is \([0,2]\) (see Figure \( \PageIndex{14} \)).

   
   Figure \( \PageIndex{14} \)

   Graphically, to check \(2x-x^2 \geq |x-1|-1\), we set \(h(x) = 2x-x^2\) and \(i(x) = |x-1|-1\) and look for the \(x\) values where the graph of \(h\) is above the the graph of \(i\) (the solution of \(h(x) > i(x)\)) as well as the \(x\)-coordinates of the intersection points of both graphs (where \(h(x)=i(x)\)). The combined sign chart is given in Figure \( \PageIndex{15} \).

   
   Figure \( \PageIndex{15} \)

Subsection Footnotes

³ This fun fact will be explored in Chapter 3 as the Intermediate Value Theorem. It has significant importance in Calculus.

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Applications Involving Quadratic Inequalities

One of the classic applications of inequalities is the notion of tolerances. The underlying concept of Calculus can be phrased in terms of tolerances, so this is well worth your attention.

Recall that for real numbers \(x\) and \(c\), the quantity \(|x-c|\) may be interpreted as the distance from \(x\) to \(c\). Solving inequalities of the form \(|x-c| \leq d\) for \(d \geq 0\) can then be interpreted as finding all numbers \(x\) which lie within \(d\) units of \(c\). We can think of the number \(d\) as a "tolerance" and our solutions \(x\) as being within an accepted tolerance of \(c\). We use this principle in the next example.

Example \( \PageIndex{5} \)

The area \(A\) (in square inches) of a square piece of particle board which measures \(x\) inches on each side is \(A(x) = x^2\). Suppose a manufacturer needs to produce a \(24\) inch by \(24\) inch square piece of particle board as part of a home office desk kit. How close does the side of the piece of particle board need to be cut to \(24\) inches to guarantee that the area of the piece is within a tolerance of \(0.25\) square inches of the target area of \(576\) square inches?

Solution

Mathematically, we express the desire for the area \(A(x)\) to be within \(0.25\) square inches of \(576\) as \[|A - 576| \leq 0.25. \nonumber \]Since \(A(x) = x^2\), we get \(|x^2 - 576| \leq 0.25\), which is equivalent to \(-0.25 \leq x^2 - 576 \leq 0.25\).

One way to proceed at this point is to solve the two inequalities \(-0.25 \leq x^2 - 576\) and \(x^2 - 576 \leq 0.25\) individually using sign diagrams and then taking the intersection of the solution sets. While this way will (eventually) lead to the correct answer, we take this opportunity to showcase the increasing property of the square root:

If \(0 \leq a \leq b\), then \(\sqrt{a} \leq \sqrt{b}\).

To use this property, we proceed as follows

\[\begin{array}{crcl}
& -0.25 \leq & x^2 - 576 & \leq 0.25 \\
\implies & 575.75 \leq & x^2 & \leq 576.25 \\
\implies & \sqrt{575.75} \leq & \sqrt{x^2} & \leq \sqrt{576.25} \\
\implies & \sqrt{575.75} \leq & |x| & \leq \sqrt{576.25} \\
\end{array} \nonumber \]

We find the solution to \(\sqrt{575.75} \leq |x|\) to be \(\left(-\infty, -\sqrt{575.75} \, \right] \cup \left[\sqrt{575.75}, \infty \right)\) and the solution to \(|x| \leq \sqrt{576.25}\) to be \(\left[-\sqrt{576.25}, \sqrt{576.25} \, \right]\). To solve \(\sqrt{575.75} \leq |x| \leq \sqrt{576.25}\), we intersect these two sets to get \([-\sqrt{576.25}, -\sqrt{575.75}] \cup [\sqrt{575.75},\sqrt{576.25}]\). Since \(x\) represents a length, we discard the negative answers and get \([\sqrt{575.75},\sqrt{576.25}]\). This means that the side of the piece of particle board must be cut between \(\sqrt{575.75} \approx 23.995\) and \(\sqrt{576.25} \approx 24.005\) inches, a tolerance of (approximately) \(0.005\) inches of the target length of \(24\) inches.

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Regions in Planes

Our last example in the section demonstrates how inequalities can be used to describe regions in the plane, as we saw earlier in Section 1.2.

Example \( \PageIndex{6} \)

Sketch the following relations.

1. \(R = \{ (x,y) : y > |x| \}\)
2. \(S = \{ (x,y) : y \leq 2-x^2 \}\)
3. \(T = \{ (x,y) : |x| < y \leq 2-x^2 \}\)

Solution

1. The relation \(R\) consists of all points \((x,y)\) whose \(y\)-coordinate is greater than \(|x|\). If we graph \(y=|x|\), then we want all of the points in the plane above the points on the graph. Dotting the graph of \(y=|x|\) as we have done before to indicate that the points on the graph itself are not in the relation, we get the shaded region in Figure \( \PageIndex{16}A \).

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2. For a point to be in \(S\), its \(y\)-coordinate must be less than or equal to the \(y\)-coordinate on the parabola \(y=2-x^2\). This is the set of all points below or on the parabola \(y=2-x^2\). See Figure \( \PageIndex{16}B \).

   
   Figures \( \PageIndex{16}A \) (left) and \( \PageIndex{16}B \) (right)

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3. Finally, the relation \(T\) takes the points whose \(y\)-coordinates satisfy both the conditions given in \(R\) and those of \(S\). Thus we shade the region between \(y=|x|\) and \(y=2-x^2\), keeping those points on the parabola, but not the points on \(y=|x|\). To get an accurate graph, we need to find where these two graphs intersect, so we set \(|x| = 2-x^2\). Proceeding as before, breaking this equation into cases, we get \(x=-1,1\). Graphing yields Figure \( \PageIndex{17} \).

   
   Figure \( \PageIndex{17} \)