2.4 Inequalities with Absolute Value and Quadratic Functions

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

Example 1

Let $f (x) = 2 x - 1$ and $g ⁢ open parenthesis x close parenthesis equals 5$ .

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

To solve $f (x) = g (x)$ , we replace $f (x)$ with $2 ⁢ x minus 1$ and $g (x)$ with $5$ to get $2 x - 1 = 5$ . Solving for $x$ , we get $x equals 3$ .
The inequality $f (x) < g (x)$ is equivalent to $2 ⁢ x minus 1 less than 5$ . Solving gives $x less than 3$ or $open parenthesis minus ∞, 3 close parenthesis$ .
To find where $f (x) > g (x)$ , we solve $2 ⁢ x minus 1 greater than 5$ . We get $x greater than 3$ , or $open parenthesis 3, ∞ close parenthesis$ .
To graph $y = f (x)$ , we graph $y = 2 x - 1$ , which is a line with a $y$ -intercept of $open parenthesis 0, minus 1 close parenthesis$ and a slope of $2$ . The graph of $y = g (x)$ is $y equals 5$ which is a horizontal line through $open parenthesis 0, 5 close parenthesis$ .
Figure 2.76
To see the connection between the graph and the Algebra, we recall the Fundamental Graphing Principle for Functions in Section: 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 1, we found $x equals 3$ is the solution to $f (x) = g (x)$ . Sure enough, $f ⁢ open parenthesis 3 close parenthesis equals 5$ and $g ⁢ open parenthesis 3 close parenthesis equals 5$ so that the point $open parenthesis 3, 5 close parenthesis$ is on both graphs. In other words, the graphs of $f$ and $g$ intersect at $open parenthesis 3, 5 close parenthesis$ . In part 2, we set $f (x) < g (x)$ and solved to find $x less than 3$ . For $x less than 3$ , 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 3, we solved $f (x) > g (x)$ and found $x greater than 3$ . For $x greater than 3$ , 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 $x$ .
Figure 2.77 $f (x) < g (x)$ on $open parenthesis minus ∞, 3 close parenthesis$
Figure 2.78 $f (x) > g (x)$ on $open parenthesis 3, ∞ close parenthesis$

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

Graphical Interpretation of 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 $y = g (x)$ .

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

Example 2

The graphs of $f$ and $g$ are below. (The graph of $y = g (x)$ is bolded.) Use these graphs to answer the following questions.

Figure 2.79

Solve $f (x) = g (x)$ .
Solve $f (x) < g (x)$ .
Solve $f (x) \geq g (x)$ .

To solve $f (x) = g (x)$ , we look for where the graphs of $f$ and $g$ intersect. These appear to be at the points $open parenthesis minus 1, 2 close parenthesis$ and $open parenthesis 1, 2 close parenthesis$ , so our solutions to $f (x) = g (x)$ are $x = - 1$ and $x equals 1$ .
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 $minus 1$ and greater than $1$ . Our solution is $(- \infty, - 1) \cup (1, \infty)$ .
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 $x equals 1$ , on the interval $open parenthesis minus 1, 1 close parenthesis$ . Hence, our solution to $f (x) \geq g (x)$ is $open bracket minus 1, 1 close bracket$ .
Figure 2.80 $f (x) < g (x)$
Figure 2.81 $f (x) \geq g (x)$

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

Theorem

Inequalities Involving the Absolute Value: Let $c$ be a real number.

For $c greater than 0$ , $| x | < c$ is equivalent to $- c < x < c$ .
For $c greater than 0$ , $| x | \leq c$ is equivalent to $- c \leq x \leq c$ .
For $c less than or equal to 0$ , $| x | < c$ has no solution, and for $c less than 0$ , $| x | \leq c$ has no solution.
For $c greater than or equal to 0$ , $| x | > c$ is equivalent to $x < - c$ or $x > c$ .
For $c greater than or equal to 0$ , $| x | \geq c$ is equivalent to $x \leq - c$ or $x \geq c$ .
For $c less than 0$ , $| x | > c$ and $| x | \geq c$ are true for all real numbers.

As with Theorem in Section, we could argue Theorem using cases. However, in light of what we have developed in this section, we can understand these statements graphically. For instance, if $c greater than 0$ , the graph of $y = c$ is a horizontal line which lies above the $x$ -axis through $open parenthesis 0, c close parenthesis$ . 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, we know happens when $x = c$ or $x = - c$ . Graphing, we get

Coordinate-plane figure. — **Figure 2.82**

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 can be shown similarly.

Example 3

Solve the following inequalities analytically; check your answers graphically.

$vertical bar x minus 1 vertical bar greater than or equal to 3$
$4 - 3 | 2 x + 1 | > - 2$
$2 less than vertical bar x minus 1 vertical bar less than or equal to 5$
$| x + 1 | \geq \frac{x + 4}{2}$

From Theorem, $vertical bar x minus 1 vertical bar greater than or equal to 3$ is equivalent to $x - 1 \leq - 3$ or $x minus 1 greater than or equal to 3$ . Solving, we get $x less than or equal to minus 2$ or $x greater than or equal to 4$ , which, in interval notation is $(- \infty, - 2] \cup [4, \infty)$ . Graphically, we have
Figure 2.83
We see that the graph of $y = | x - 1 |$ is above the horizontal line $y equals 3$ for $x less than minus 2$ and $x greater than 4$ hence this is where $vertical bar x minus 1 vertical bar greater than 3$ . The two graphs intersect when $x = - 2$ and $x equals 4$ , so we have graphical confirmation of our analytic solution.
To solve $4 - 3 | 2 x + 1 | > - 2$ analytically, we first isolate the absolute value before applying Theorem. To that end, we get $- 3 | 2 x + 1 | > - 6$ or $vertical bar 2 ⁢ x plus 1 vertical bar less than 2$ . Rewriting, we now have $- 2 < 2 x + 1 < 2$ so that $- \frac{3}{2} < x < \frac{1}{2}$ . In interval notation, we write $(- \frac{3}{2}, \frac{1}{2})$ . Graphically we see that the graph of $y = 4 - 3 | 2 x + 1 |$ is above $y = - 2$ for $x$ values between $- \frac{3}{2}$ and $\frac{1}{2}$ .
Figure 2.84
Rewriting the compound inequality $2 less than vertical bar x minus 1 vertical bar less than or equal to 5$ as ` $2 less than vertical bar x minus 1 vertical bar$ and $vertical bar x minus 1 vertical bar less than or equal to 5$ ' allows us to solve each piece using Theorem. The first inequality, $2 less than vertical bar x minus 1 vertical bar$ can be re-written as $vertical bar x minus 1 vertical bar greater than 2$ so $x - 1 < - 2$ or $x minus 1 greater than 2$ . We get $x less than minus 1$ or $x greater than 3$ . Our solution to the first inequality is then $(- \infty, - 1) \cup (3, \infty)$ . For $vertical bar x minus 1 vertical bar less than or equal to 5$ , we combine results in Theorems and to get $- 5 \leq x - 1 \leq 5$ so that $minus 4 less than or equal to x less than or equal to 6$ , or $open bracket minus 4, 6 close bracket$ . Our solution to $2 less than vertical bar x minus 1 vertical bar less than or equal to 5$ is comprised of values of $x$ which satisfy both parts of the inequality, so we take the intersection¹ of $(- \infty, - 1) \cup (3, \infty)$ and $open bracket minus 4, 6 close bracket$ to get $[- 4, - 1) \cup (3, 6]$ . Graphically, we see that the graph of $y = | x - 1 |$ is `between' the horizontal lines $y equals 2$ and $y equals 5$ for $x$ values between $minus 4$ and $minus 1$ as well as those between $3$ and $6$ . Including the $x$ values where $y = | x - 1 |$ and $y equals 5$ intersect, we get
Figure 2.85
We need to exercise some special caution when solving $| x + 1 | \geq \frac{x + 4}{2}$ . As we saw in Example in Section, when variables are both inside and outside of the absolute value, it's usually best to refer to the definition of absolute value, Definition, to remove the absolute values and proceed from there. To that end, we have $| x + 1 | = - (x + 1)$ if $x less than minus 1$ and $| x + 1 | = x + 1$ if $x greater than or equal to minus 1$ . We break the inequality into cases, the first case being when $x less than minus 1$ . For these values of $x$ , our inequality becomes $- (x + 1) \geq \frac{x + 4}{2}$ . Solving, we get $- 2 x - 2 \geq x + 4$ , so that $minus 3 ⁢ x greater than or equal to 6$ , which means $x less than or equal to minus 2$ . Since all of these solutions fall into the category $x less than minus 1$ , we keep them all. For the second case, we assume $x greater than or equal to minus 1$ . Our inequality becomes $x + 1 \geq \frac{x + 4}{2}$ , which gives $2 x + 2 \geq x + 4$ or $x greater than or equal to 2$ . Since all of these values of $x$ are greater than or equal to $minus 1$ , we accept all of these solutions as well. Our final answer is $(- \infty, - 2] \cup [2, \infty)$ .
Figure 2.86

We now turn our attention to quadratic inequalities. In the last example of Section, we needed to determine the solution to $x^{2} - x - 6 < 0$ . We will now re-visit this problem using some of the techniques developed in this section not only to reinforce our solution in Section, but to also help formulate a general analytic procedure for solving all quadratic inequalities. If we consider $f (x) = x^{2} - x - 6$ and $g ⁢ open parenthesis x close parenthesis equals 0$ , 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.

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 $x equals 3$ in this case and they divide the domain (the $x$ -axis) into three intervals: $(- \infty, - 2)$ , $open parenthesis minus 2, 3 close parenthesis$ and $open parenthesis 3, ∞ close parenthesis$ . For every number in $(- \infty, - 2)$ , the graph of $f$ is above the $x$ -axis; in other words, $f ⁢ open parenthesis x close parenthesis greater than 0$ for all $x$ in $(- \infty, - 2)$ . Similarly, $f ⁢ open parenthesis x close parenthesis less than 0$ for all $x$ in $open parenthesis minus 2, 3 close parenthesis$ , and $f ⁢ open parenthesis x close parenthesis greater than 0$ for all $x$ in $open parenthesis 3, ∞ close parenthesis$ . We can schematically represent this with the sign diagram below.

Here, the $(+)$ above a portion of the number line indicates $f ⁢ open parenthesis x close parenthesis greater than 0$ for those values of $x$ ; the $(-)$ indicates $f ⁢ open parenthesis x close parenthesis less than 0$ 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 $f ⁢ open parenthesis x close parenthesis less than 0$ is $open parenthesis minus 2, 3 close parenthesis$ .

Our next goal is to establish a procedure by which we can generate the sign diagram without graphing the function. An important property² 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. 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. This gives us the following.

Steps for Solving a Quadratic Inequality

Rewrite the inequality, if necessary, as a quadratic function $f (x)$ on one side of the inequality and $0$ on the other.
Find the zeros of $f$ and place them on the number line with the number $0$ above them.
Choose a real number, called a test value, in each of the intervals determined in step 2.
Determine the sign of $f (x)$ for each test value in step 3, and write that sign above the corresponding interval.
Choose the intervals which correspond to the correct sign to solve the inequality.

Example 4

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

$2 ⁢ x squared less than or equal to 3 minus x$
$x squared minus 2 ⁢ x greater than 1$
$x squared plus 1 less than or equal to 2 ⁢ x$
$2 x - x^{2} \geq | x - 1 | - 1$

To solve $2 ⁢ x squared less than or equal to 3 minus x$ , we first get $0$ on one side of the inequality which yields $2 x^{2} + x - 3 \leq 0$ . We find the zeros of $f (x) = 2 x^{2} + x - 3$ by solving $2 x^{2} + x - 3 = 0$ for $x$ . Factoring gives $(2 x + 3) (x - 1) = 0$ , so $x = - \frac{3}{2}$ or $x equals 1$ . We place these values on the number line with $0$ above them and choose test values in the intervals $(- \infty, - \frac{3}{2})$ , $(- \frac{3}{2}, 1)$ and $open parenthesis 1, ∞ close parenthesis$ . For the interval $(- \infty, - \frac{3}{2})$ , we choose³ $x = - 2$ ; for $(- \frac{3}{2}, 1)$ , we pick $x equals 0$ ; and for $open parenthesis 1, ∞ close parenthesis$ , $x equals 2$ . Evaluating the function at the three test values gives us $f (- 2) = 3 > 0$ , so we place $(+)$ above $(- \infty, - \frac{3}{2})$ ; $f (0) = - 3 < 0$ , so $(-)$ goes above the interval $(- \frac{3}{2}, 1)$ ; and, $f ⁢ open parenthesis 2 close parenthesis equals 7$ , which means $(+)$ is placed above $open parenthesis 1, ∞ close parenthesis$ . Since we are solving $2 x^{2} + x - 3 \leq 0$ , we look for solutions to $2 x^{2} + x - 3 < 0$ as well as solutions for $2 x^{2} + x - 3 = 0$ . For $2 x^{2} + x - 3 < 0$ , we need the intervals which we have a $(-)$ . Checking the sign diagram, we see this is $(- \frac{3}{2}, 1)$ . We know $2 x^{2} + x - 3 = 0$ when $x = - \frac{3}{2}$ and $x equals 1$ , so our final answer is $[- \frac{3}{2}, 1]$ .
To verify our solution graphically, we refer to the original inequality, $2 ⁢ x squared less than or equal to 3 minus x$ . We let $g ⁢ open parenthesis x close parenthesis equals 2 ⁢ x squared$ 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 $g (x) = h (x)$ ). The graphs of $g$ and $h$ are given on the right with the sign chart on the left.
Figure 2.89
Figure 2.90
Once again, we re-write $x squared minus 2 ⁢ x greater than 1$ as $x^{2} - 2 x - 1 > 0$ and we identify $f (x) = x^{2} - 2 x - 1$ . When we go to find the zeros of $f$ , we find, to our chagrin, that the quadratic $x^{2} - 2 x - 1$ doesn't factor nicely. Hence, we resort to the quadratic formula to solve $x^{2} - 2 x - 1 = 0$ , 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 $x = - 1$ , $x equals 0$ and $x equals 3$ as our test values and find $f (- 1) = 2$ , which is $(+)$ ; $f (0) = - 1$ which is $(-)$ ; and $f ⁢ open parenthesis 3 close parenthesis equals 2$ which is $(+)$ again. Our solution to $x^{2} - 2 x - 1 > 0$ is where we have $(+)$ , so, in interval notation $(- \infty, 1 - \sqrt{2}) \cup (1 + \sqrt{2}, \infty)$ . To check the inequality $x squared minus 2 ⁢ x greater than 1$ graphically, we set $g (x) = x^{2} - 2 x$ and $h ⁢ open parenthesis x close parenthesis equals 1$ . We are looking for the $x$ values where the graph of $g$ is above the graph of $h$ . As before we present the graphs on the right and the sign chart on the left.
Figure 2.91
Figure 2.92
To solve $x squared plus 1 less than or equal to 2 ⁢ x$ , as before, we solve $x^{2} - 2 x + 1 \leq 0$ . Setting $f (x) = x^{2} - 2 x + 1 = 0$ , we find the only one zero of $f$ , $x equals 1$ . This one $x$ value divides the number line into two intervals, from which we choose $x equals 0$ and $x equals 2$ as test values. We find $f ⁢ open parenthesis 0 close parenthesis equals 1 greater than 0$ and $f ⁢ open parenthesis 2 close parenthesis equals 1 greater than 0$ . Since we are looking for solutions to $x^{2} - 2 x + 1 \leq 0$ , we are looking for $x$ values where $x^{2} - 2 x + 1 < 0$ as well as where $x^{2} - 2 x + 1 = 0$ . Looking at our sign diagram, there are no places where $x^{2} - 2 x + 1 < 0$ (there are no $(-)$ ), so our solution is only $x equals 1$ (where $x^{2} - 2 x + 1 = 0$ ). We write this as $open brace 1 close brace$ . Graphically, we solve $x squared plus 1 less than or equal to 2 ⁢ x$ by graphing $g (x) = x^{2} + 1$ and $h ⁢ open parenthesis x close parenthesis equals 2 ⁢ x$ . We are looking for the $x$ values where the graph of $g$ is below the graph of $h$ (for $x squared plus 1 less than 2 ⁢ x$ ) and where the two graphs intersect ( $x^{2} + 1 = 2 x$ ). Notice that the line and the parabola touch at $open parenthesis 1, 2 close parenthesis$ , but the parabola is always above the line otherwise.⁴
Figure 2.93
Figure 2.94
To solve our last inequality, $2 x - x^{2} \geq | x - 1 | - 1$ , we re-write the absolute value using cases. For $x less than 1$ , $| x - 1 | = - (x - 1) = 1 - x$ , so we get $2 x - x^{2} \geq 1 - x - 1$ , or $x squared minus 3 ⁢ x less than or equal to 0$ . Finding the zeros of $f (x) = x^{2} - 3 x$ , we get $x equals 0$ and $x equals 3$ . However, we are only concerned with the portion of the number line where $x less than 1$ , so the only zero that we concern ourselves with is $x equals 0$ . This divides the interval $x less than 1$ into two intervals: $open parenthesis minus ∞, 0 close parenthesis$ and $open parenthesis 0, 1 close parenthesis$ . We choose $x = - 1$ and $x = \frac{1}{2}$ as our test values. We find $f (- 1) = 4$ and $f (\frac{1}{2}) = - \frac{5}{4}$ . Hence, our solution to $x squared minus 3 ⁢ x less than or equal to 0$ for $x less than 1$ is $open bracket 0, 1 close parenthesis$ . Next, we turn our attention to the case $x greater than or equal to 1$ . Here, $| x - 1 | = x - 1$ , so our original inequality becomes $2 x - x^{2} \geq x - 1 - 1$ , 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 $x equals 2$ . Of these, only $x equals 2$ lies in the region $x greater than or equal to 1$ , so we ignore $x = - 1$ . Our test intervals are now $open bracket 1, 2 close parenthesis$ and $open parenthesis 2, ∞ close parenthesis$ . We choose $x equals 1$ and $x equals 3$ as our test values and find $g (1) = - 2$ and $g ⁢ open parenthesis 3 close parenthesis equals 4$ . Hence, our solution to $g (x) = x^{2} - x - 2 \leq 0$ , in this region is $open bracket 1, 2 close parenthesis$ .
Figure 2.95
Figure 2.96
Combining these into one sign diagram, we have that our solution is $open bracket 0, 2 close bracket$ . Graphically, to check $2 x - x^{2} \geq | x - 1 | - 1$ , we set $h (x) = 2 x - x^{2}$ 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 on the left and the graphs are on the right.
Figure 2.97
Figure 2.98

One of the classic applications of inequalities is the notion of tolerances.⁵ 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 $d greater than or equal to 0$ 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 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?

Mathematically, we express the desire for the area $A (x)$ to be within $0.25$ square inches of $576$ as $vertical bar A minus 576 vertical bar less than or equal to 0.25$ . Since $A (x) = x^{2}$ , we get $vertical bar x squared minus 576 vertical bar less than or equal to 0.25$ , 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 $x squared minus 576 less than or equal to 0.25$ 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 $0 less than or equal to a less than or equal to b$ , then $\sqrt{a} \leq \sqrt{b}$ . To use this property, we proceed as follows

$\begin{matrix} - 0.25 \leq & x^{2} - 576 & \leq 0.25 \\ 575.75 \leq & x^{2} & \leq 576.25 & (add 576 across the inequalities.) \\ \sqrt{575.75} \leq & \sqrt{x^{2}} & \leq \sqrt{576.25} & (take square roots.) \\ \sqrt{575.75} \leq & | x | & \leq \sqrt{576.25} & (\sqrt{x^{2}} = | x |) \end{matrix}$

By Theorem, we find the solution to $the square root of 575.75 less than or equal to vertical bar x vertical bar$ to be $(- \infty, - \sqrt{575.75}] \cup [\sqrt{575.75}, \infty)$ and the solution to $vertical bar x vertical bar less than or equal to the square root of 576.25$ to be $[- \sqrt{576.25}, \sqrt{576.25}]$ . 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.

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

Example 6

Sketch the following relations.

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

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 below on the left.
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 $y = 2 - x^{2}$ .
Figure 2.99 The graph of $R$
Figure 2.100 The graph of $S$
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 2.101 The graph of $T$

Exercises

In Exercises -, solve the inequality. Write your answer using interval notation.

$vertical bar 3 ⁢ x minus 5 vertical bar less than or equal to 4$
$vertical bar 7 ⁢ x plus 2 vertical bar greater than 10$
$| 2 x + 1 | - 5 < 0$
$| 2 - x | - 4 \geq - 3$
$| 3 x + 5 | + 2 < 1$
$2 | 7 - x | + 4 > 1$
$2 less than or equal to vertical bar 4 minus x vertical bar less than 7$
$1 less than vertical bar 2 ⁢ x minus 9 vertical bar less than or equal to 3$
$| x + 3 | \geq | 6 x + 9 |$
$| x - 3 | - | 2 x + 1 | < 0$
$| 1 - 2 x | \geq x + 5$
$x + 5 < | x + 5 |$
$x greater than or equal to vertical bar x plus 1 vertical bar$
$| 2 x + 1 | \leq 6 - x$
$x + | 2 x - 3 | < 2$
$| 3 - x | \geq x - 5$
$x^{2} + 2 x - 3 \geq 0$
$16 x^{2} + 8 x + 1 > 0$
$x squared plus 9 less than 6 ⁢ x$
$9 ⁢ x squared plus 16 greater than or equal to 24 ⁢ x$
$x squared plus 4 less than or equal to 4 ⁢ x$
$x squared plus 1 less than 0$
$3 ⁢ x squared less than or equal to 11 ⁢ x plus 4$
$x > x^{2}$
$2 x^{2} - 4 x - 1 > 0$
$5 ⁢ x plus 4 less than or equal to 3 ⁢ x squared$
$2 less than or equal to vertical bar x squared minus 9 vertical bar less than 9$
$x squared less than or equal to vertical bar 4 ⁢ x minus 3 vertical bar$
$x^{2} + x + 1 \geq 0$
$x^{2} \geq | x |$
$x | x + 5 | \geq - 6$
$x ⁢ vertical bar x minus 3 vertical bar less than 2$
The profit, in dollars, made by selling $x$ bottles of $100 %$ All-Natural Certified Free-Trade Organic Sasquatch Tonic is given by $P (x) = - x^{2} + 25 x - 100$ , for $0 less than or equal to x less than or equal to 35$ . How many bottles of tonic must be sold to make at least $$ 50$ in profit?
Suppose $C (x) = x^{2} - 10 x + 27$ , $x greater than or equal to 0$ represents the costs, in hundreds of dollars, to produce $x$ thousand pens. Find the number of pens which can be produced for no more than $$ 1100$ .
The temperature $T$ , in degrees Fahrenheit, $t$ hours after 6 AM is given by $T (t) = - \frac{1}{2} t^{2} + 8 t + 32$ , for $0 less than or equal to t less than or equal to 12$ . When is it warmer than $42 degrees$ Fahrenheit?
The height $h$ in feet of a model rocket above the ground $t$ seconds after lift-off is given by $h (t) = - 5 t^{2} + 100 t$ , for $0 less than or equal to t less than or equal to 20$ . When is the rocket at least $250$ feet off the ground? Round your answer to two decimal places.
If a slingshot is used to shoot a marble straight up into the air from 2 meters above the ground with an initial velocity of 30 meters per second, for what values of time $t$ will the marble be over 35 meters above the ground? (Refer to Exercise in Section for assistance if needed.) Round your answers to two decimal places.
What temperature values in degrees Celsius are equivalent to the temperature range $50 degrees ⁢ F$ to $95 degrees ⁢ F$ ? (Refer to Exercise in Section for assistance if needed.)
Find all real numbers $x$ so that $x$ is within $4$ units of $2$ .
Find all real numbers $x$ so that $3 ⁢ x$ is within $2$ units of $minus 1$ .
Find all real numbers $x$ so that $x^{2}$ is within $1$ unit of $3$ .
Find all real numbers $x$ so that $x^{2}$ is at least $7$ units away from $4$ .
The surface area $S$ of a cube with edge length $x$ is given by $S ⁢ open parenthesis x close parenthesis equals 6 ⁢ x squared$ for $x greater than 0$ . Suppose the cubes your company manufactures are supposed to have a surface area of exactly 42 square centimeters, but the machines you own are old and cannot always make a cube with the precise surface area desired. Write an inequality using absolute value that says the surface area of a given cube is no more than 3 square centimeters away (high or low) from the target of 42 square centimeters. Solve the inequality and write your answer using interval notation.
Suppose $f$ is a function, $L$ is a real number and $ε$ is a positive number. Discuss with your classmates what the inequality $| f (x) - L | < ε$ means algebraically and graphically.⁶
$R = {(x, y) : y \leq x - 1}$
$R = {(x, y) : y > x^{2} + 1}$
$R = {(x, y) : - 1 < y \leq 2 x + 1}$
$R = {(x, y) : x^{2} \leq y < x + 2}$
$R = {(x, y) : | x | - 4 < y < 2 - x}$
$R = {(x, y) : x^{2} < y \leq | 4 x - 3 |}$
Prove the second, third and fourth parts of Theorem.

In Exercises -, write and solve an inequality involving absolute values for the given statement.

In Exercises -, sketch the graph of the relation.

Answers

$[\frac{1}{3}, 3]$
$(- \infty, - \frac{12}{7}) \cup (\frac{8}{7}, \infty)$
$open parenthesis minus 3, 2 close parenthesis$
$open parenthesis minus ∞, 1 close bracket union open bracket 3, ∞ close parenthesis$
No solution
$open parenthesis minus ∞, ∞ close parenthesis$
$open parenthesis minus 3, 2 close bracket union open bracket 6, 11 close parenthesis$
$open bracket 3, 4 close parenthesis union open parenthesis 5, 6 close bracket$
$[- \frac{12}{7}, - \frac{6}{5}]$
$(- \infty, - 4) \cup (\frac{2}{3}, \infty)$
$(- \infty, - \frac{4}{3}] \cup [6, \infty)$
$(- \infty, - 5)$
No Solution.
$[- 7, \frac{5}{3}]$
$(1, \frac{5}{3})$
$open parenthesis minus ∞, ∞ close parenthesis$
$(- \infty, - 3] \cup [1, \infty)$
$(- \infty, - \frac{1}{4}) \cup (- \frac{1}{4}, \infty)$
No solution
$open parenthesis minus ∞, ∞ close parenthesis$
$open brace 2 close brace$
No solution
$[- \frac{1}{3}, 4]$
$open parenthesis 0, 1 close parenthesis$
$(- \infty, 1 - \frac{\sqrt{6}}{2}) \cup (1 + \frac{\sqrt{6}}{2}, \infty)$
$(- \infty, \frac{5 - \sqrt{73}}{6}] \cup [\frac{5 + \sqrt{73}}{6}, \infty)$
$(- 3 \sqrt{2}, - \sqrt{11}] \cup [- \sqrt{7}, 0) \cup (0, \sqrt{7}] \cup [\sqrt{11}, 3 \sqrt{2})$
$[- 2 - \sqrt{7}, - 2 + \sqrt{7}] \cup [1, 3]$
$open parenthesis minus ∞, ∞ close parenthesis$
$(- \infty, - 1] \cup {0} \cup [1, \infty)$
$[- 6, - 3] \cup [- 2, \infty)$
$(- \infty, 1) \cup (2, \frac{3 + \sqrt{17}}{2})$
$P ⁢ open parenthesis x close parenthesis greater than or equal to 50$ on $open bracket 10, 15 close bracket$ . This means anywhere between 10 and 15 bottles of tonic need to be sold to earn at least $$ 50$ in profit.
$C ⁢ open parenthesis x close parenthesis less than or equal to 11$ on $open bracket 2, 8 close bracket$ . This means anywhere between 2000 and 8000 pens can be produced and the cost will not exceed $$ 1100$ .
$T ⁢ open parenthesis t close parenthesis greater than 42$ on $(8 - 2 \sqrt{11}, 8 + 2 \sqrt{11}) \approx (1.37, 14.63)$ , which corresponds to between 7:22 AM (1.37 hours after 6 AM) to 8:38 PM (14.63 hours after 6 AM.) However, since the model is valid only for $t$ , $0 less than or equal to t less than or equal to 12$ , we restrict our answer and find it is warmer than $42 degrees$ Fahrenheit from 7:22 AM to 6 PM.
$h ⁢ open parenthesis t close parenthesis greater than or equal to 250$ on $[10 - 5 \sqrt{2}, 10 + 5 \sqrt{2}] \approx [2.93, 17.07]$ . This means the rocket is at least 250 feet off the ground between 2.93 and 17.07 seconds after lift off.
$s (t) = - 4.9 t^{2} + 30 t + 2$ . $s ⁢ open parenthesis t close parenthesis greater than 35$ on (approximately) $open parenthesis 1.44, 4.68 close parenthesis$ . This means between 1.44 and 4.68 seconds after it is launched into the air, the marble is more than 35 feet off the ground.
From our previous work $C (F) = \frac{5}{9} (F - 32)$ so $50 less than or equal to F less than or equal to 95$ becomes $10 less than or equal to C less than or equal to 35$ .
$vertical bar x minus 2 vertical bar less than or equal to 4$ , $open bracket minus 2, 6 close bracket$
$vertical bar 3 ⁢ x plus 1 vertical bar less than or equal to 2$ , $[- 1, \frac{1}{3}]$
$vertical bar x squared minus 3 vertical bar less than or equal to 1$ , $[- 2, - \sqrt{2}] \cup [\sqrt{2}, 2]$
$vertical bar x squared minus 4 vertical bar greater than or equal to 7$ , $(- \infty, - \sqrt{11}] \cup [\sqrt{11}, \infty)$
Solving $vertical bar S ⁢ open parenthesis x close parenthesis minus 42 vertical bar less than or equal to 3$ , and disregarding the negative solutions yields $[\sqrt{\frac{13}{2}}, \sqrt{\frac{15}{2}}] \approx [2.550, 2.739]$ . The edge length must be within $2.550$ and $2.739$ centimeters.
Figure 2.102
Figure 2.103
Figure 2.104
Figure 2.105
Figure 2.106
Figure 2.107

See Definition in Section . ↩
We will give this property a name in Chapter and revisit this concept then. ↩
We have to choose something in each interval. If you don't like our choices, please feel free to choose different numbers. You'll get the same sign chart. ↩
In this case, we say the line $y equals 2 ⁢ x$ is tangent to $y = x^{2} + 1$ at $open parenthesis 1, 2 close parenthesis$ . Finding tangent lines to arbitrary functions is a fundamental problem solved, in general, with Calculus. ↩
The underlying concept of Calculus can be phrased in terms of tolerances, so this is well worth your attention. ↩
Understanding this type of inequality is really important in Calculus. ↩

Adapted from Precalculus, 3rd corrected edition, by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.