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Precalculus Study Guide

Chapter 1: Solving Inequalities

Introduction

An equation is solved by performing equivalent operations on both sides, ultimately resulting in the unknown in the equation being isolated on one side of the equation. For example, the equation

$2 x + 7 = 5 x - 4$

can be solved by subtracting $2 x$ from both sides, then adding 4 to both sides. This would result in the equation

$11 = 3 x$

from which the solution $x = \frac{11}{3}$ follows by division of both sides by 3.

An inequality such as

$2 x + 7 < 5 x - 4$

is also solved by performing equivalent operations on both sides, but with the caveat that some operations can reverse the "sense" of the inequality. This first chapter studies the various techniques that can be used to solve the types of inequalities found in the typical calculus course.

Often, the inequalities found in calculus involve $f (x) = |x|$ , the absolute value function, explained naively here, and in more detail in Chapter 10.

Chapter Glossary

abscissa

absolute value

inequality

intersection

interval
nonnegative

open interval

positive

real line

subset

union

The Absolute Value Function, $|x|$

The absolute value function, written as

$f (x) = |x|$

is always a nonnegative quantity. For example, the absolute value of $- 3$ is the positive number 3.

Viewing the change from $- 3$ to 3 as "dropping the minus sign" obscures the abstract definition of this function.

It's much better to see the transition of $- 3$ to 3 as attained by multiplying the negative number $- 3$ by $- 1$ to form +3, the additive inverse of $- 3$ . Hence, the action of the absolute value function should be seen as

$|- 3| = −1 (−3) = 3$

that is, don't drop a minus sign, but make the negative number positive by introducing a second minus sign. This is worth repeating.

Make a negative number positive by negating the negative number

Thus, the absolute value function is implemented as follows:

The absolute value of a ${\begin{array}{c} positive quantity is the quantity itself \\ negative quantity is the negative of that quantity \end{array}$

So, for example,

$|5 - 3 x|$

the absolute value of the quantity $5 - 3 x$ , is just $5 - 3 x$ for an $x$ that keeps $5 - 3 x$ positive. On the other hand, we have

$|5 - 3 x| = - (5 - 3 x) = −5 + 3 x$

for an $x$ that keeps $5 - 3 x$ negative. (Of course, if $x = \frac{5}{3}$ so that $5 - 3 x = 0$ , then $|5 - 3 x| = 0$ .)

Figure 1.0.1 provides a graph of $|5 - 3 x|$ .

To the left of $x = \frac{5}{3}$ (where $5 - 3 x$ is positive), the graph shows

$|5 - 3 x| = 5 - 3 x$

To the right of $x = \frac{5}{3}$ (where $5 - 3 x$ is negative), the graph shows

$|5 - 3 x| = - (5 - 3 x) = −5 + 3 x$

Right at $x = \frac{5}{3}$ (where $5 - 3 x = 0$ ), the graphs shows $|5 - 3 x| = 0$

Figure 1.0.1 Graph of $|5 - 3 x|$

In Chapter 10, the absolute value will be written as a piecewise-defined function.

Typical Problems

Solve each of the following inequalities.

1. $2 x - 3 < 5$

2. $|2 x - 3| < 5$

3. $|2 x - 3| < |5 x - 4|$

4. $5 \leq |6 x - 1|$

5. $\frac{2}{x} - 1 \leq 7$

6. $|\frac{2}{x} - 1| \leq 7$

7. $|\frac{2}{x} - 1| < |5 x + 8|$

8. $5 \leq |\frac{2}{x} - 1|$

9. $3 < 2 x - 3$ $< 7$

10. $3 < |\frac{2}{x} - 1|$ $< 7$

Note that Problems 1 - 8 are of the form $f (x) < g (x)$ or $f (x) \leq g (x)$ , whereas Problems 9 - 10 are of the form $a < f (x)$ $< b$ .

Maple Initializations

Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.

Solutions

Problem 1.1

1.1 - Mathematical Solution

The inequality

$2 x - 3 < 5$

can be solved much like the equation $2 x - 3 = 5$ would be solved. If 3 is added to each side of the inequality, the inequality is preserved, so the first step would be

$2 x - 3 < 5$

+ 3 = + 3

______________

$2 x < 8$

from which

$x < \frac{8}{2}$ = 4

follows. Here, division by the positive number 2 preserves the inequality, so the solution $x < 4$ is obtained.

Figure 1.1.1 Graph of $f (x) = 2 x - 3$ (in black) and $g (x) = 5$ (in red)

Figure 1.1.1 highlights the solution set $\{x < 4\}$ with a green arrow. These are the values of $x$ for which the graph of $f (x)$ lies below the graph of $g (x)$ .

1.1 - Maplet Solution

The inequality

$2 x - 3 < 5$

can be solved with Inequality Tutor #1 . (Clicking this link will launch the tutor with the solution embedded as shown in Figure 1.1.2.)

The left side of the inequality is entered as $f (x)$ ; the right, as $g (x)$ . Between these two functions there are radio buttons for selecting the appropriate form of the inequality.

Note: When entering an expression in a tutor window, use * for multiplication. Thus, for this example f(x) is entered as 2*x-3.

Clicking on the Graph button generates a graph of $f (x)$ (in black), and $g (x)$ (in red). The Plot Options button provides control over the axes on the graph.

Figure 1.1.2 Thumbnail image of Inequality Tutor #1

Graphically, the solution of the given inequality consists of those values of $x$ for which the black line (i.e., $f (x)$ ) lies beneath the red line (i.e., $g (x)$ ). Clearly, this occurs on the left of the intersection of the two lines.

Clicking on the Intersections button yields $x = 4$ , the abscissa of the point of intersection.

Clicking on the Solve button generates the solution of the inequality in the form of an interval, namely,

$(- \infty, 4)$

which is equivalent to $- \infty < x$ $< 4$ .

To launch Inequality Tutor #1, click the following link: Inequality Tutor #1

1.1 - Interactive Solution

Enter the data for the problem

•	Enter the left-hand side of the inequality.

•	Context Panel: Assign to a Name≻ $f$

•	Enter the right-hand side of the inequality.

•	Context Panel: Assign to a Name≻ $g$

Enter and solve the inequality

•	Write the inequality $f < g$ and press Enter.

•	Context Panel: Solve≻Solve

Calculate the $x$ -coordinate of the intersection

•	Write the equality $f = g$ and press Enter.

•	Context Panel: Solve≻Solve

Obtain Figure 1.1.1

•	Type $f, g$ and press Enter.

•	Context Panel: Plot Builder

Click Global Options button.
Set view for axis[1]: $0 \leq x \leq 5$

1.1 - Programmatic Solution

Assign the left-hand side of the inequality to the name $f$

>	$f ≔ 2 x - 3$

Assign the right-hand side of the inequality to the name $g$

$g ≔ 5$

Solve the inequality.

>	$solve (f < g, \{x\})$

Solve for the intersection of $f (x)$ and $g (x)$

>	$solve (f = g, x)$

Obtain Figure 1.1.1

>	$plot ([f, g], x = 0 .. 5, color = [black, red])$

Problem 1.2

1.2 - Mathematical Solution

The inequality

$|2 x - 3| < 5$

can be solved after the absolute-value function is interpreted according to its definition. Thus, where $2 x - 3$ is positive, $|2 x - 3|$ can be replaced by

$2 x - 3$ and the given inequality means

$2 x - 3 < 5$

However, where $2 x - 3$ is negative, $|2 x - 3|$ must be replaced by $- (2 x - 3)$ so that the given inequality means

$- (2 x + 3) < 5$

Now this inequality can be simplified to

$2 x + 3 > - 5$

Figure 1.2.1 Graph of $f (x) = |2 x - 3|$ (in black) and $g (x) = 5$ (in red)

by multiplying through the inequality by $- 1$ , which reverses the sense of the inequality. Therefore, the two resulting inequalities can be combined to the form

$- 5 < 2 x - 3$ $< 5$

which can be solved by first adding 3 to each of the three members of the inequality, as shown below.

$- 5 < 2 x - 3$ $< 5$

+ 3 = + 3 = + 3

________________________

$- 2 < 2 x$ $< 8$

Then, the $x$ in the "center" can be isolated by dividing each member of the inequality by 2, a positive number, which will not change the sense of the inequalities. This leads to the solution

$- 1 < x$ $< 4$

Alternatively, the solution set can be expressed as the open interval $(- 1, 4)$ .

Figure 1.2.1 highlights the solution set $\{- 1 < x < 4\}$ with green arrows. These are the values of $x$ for which the graph of $f (x)$ lies below the graph of $g (x)$ .

1.2 - Maplet Solution

The inequality

$|2 x - 3| < 5$

can be solved with Inequality Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in Figure 1.2.2.

The left side of the inequality is entered as $f (x)$ , the right, as $g (x)$ . Between these two functions there are radio buttons for selecting the appropriate form of the inequality.

Clicking on the Graph button generates a graph of $f (x)$ (in black), and $g (x)$ (in red).

Figure 1.2.2 Thumbnail image of Inequality Tutor #1

The Plot Options button provides control over the axes on the graph.

Clicking on the Intersections button yields $x = - 1, 4$ , the abscissas of the points of intersection.

Clicking on the Solve button generates the solution of the inequality as the open interval

$(- 1, 4)$

which is equivalent to $- 1 < x$ $< 4$ .

To launch Inequality Tutor #1, click the following link: Inequality Tutor #1

1.2 - Interactive Solution

Enter the data for the problem

•	Enter the left-hand side of the inequality. The absolute value template can be found in the Layout palette, or can be entered from the keyboard using the vertical stroke: \|

•	Context Panel: Assign to a Name≻ $f$

•	Enter the right-hand side of the inequality.

•	Context Panel: Assign to a Name≻ $g$

Enter and solve the inequality

•	Write the inequality $f < g$ and press Enter.

•	Context Panel: Solve≻Solve

Calculate the $x$ -coordinate of the intersection

•	Write the equality $f = g$ and press Enter.

•	Context Panel: Solve≻Solve

Obtain Figure 1.2.1

•	Type $f, g$ and press Enter.

•	Context Panel: Plot Builder

Click Global Options button.
Set view for axis[1]: $- 2 \leq x \leq 5$

1.2 - Programmatic Solution

Assign the left-hand side of the inequality to the name $f$

>	$f ≔ \|2 x - 3\|$

Assign the right-hand side of the inequality to the name $g$

$g ≔ 5$

Solve the inequality.

>	$solve (f < g, \{x\})$

Solve for the intersection of $f (x)$ and $g (x)$

>	$solve (f = g, x)$

Obtain Figure 1.2.1

>	$plot ([f, g], x = - 2 .. 5, color = [black, red])$

Problem 1.3

1.3 - Mathematical Solution

The inequality

$f (x) = |2 x - 3| < |5 x - 4| = g (x)$

can be solved analytically if the absolute value functions on both sides of the inequality are interpreted according to the definition. That means there are four cases to consider, according to whether $F (x) = 2 x - 3$ and $G (x) = 5 x - 4$ are either positive or negative.

Hence, there are four cases to consider, namely,

	$F (x) = 2 x - 3$	$G (x) = 5 x - 4$
Case 1	Positive	Positive
Case 2	Positive	Negative
Case 3	Negative	Positive
Case 4	Negative	Negative

Figure 1.3.1 Graph of $|2 x - 3|$ (in black) and $|5 x - 4|$ (in red)

Case 1: $F (x)$ Positive, $G (x)$ Positive

The $x$ 's for which $F (x) = 2 x - 3$ is positive satisfy $x > 3 / 2$ and lie in the interval $(3 / 2, \infty)$ .

The $x$ 's for which $G (x) = 5 x - 4$ is positive satisfy $x > 4 / 5$ and lie in the interval $(4 / 5, \infty)$ .

The $x$ 's for which both $F (x)$ and $G (x)$ are positive lie in the intersection of these two intervals, and hence lie in $(3 / 2, \infty)$ .

If both $F (x)$ and $G (x)$ are positive, then the inequality

$|2 x - 3| < |5 x - 4|$

becomes

$2 x - 3 < 5 x - 4$

the solution of which can be obtained by subtracting $2 x$ from each side, and adding 4 to each side. This results in

$2 x - 3 < 5 x - 4$

$- 2 x + 4 = - 2 x + 4$

__________________

$1 < 3 x$

so that $x > 1 / 3$ . But not all such $x$ are valid. Only those in the interval $(3 / 2, \infty)$ can be considered. Hence, for the case when both $F (x)$ and $G (x)$ are positive, the solution of the inequality

$|2 x - 3| < |5 x - 4|$

consists of those $x$ 's in the interval $(3 / 2, \infty)$ .

Case 2: $F (x)$ Positive, $G (x)$ Negative

The $x$ 's for which $F (x) = 2 x - 3$ is positive satisfy $x > 3 / 2$ and lie in the interval $(3 / 2, \infty)$ .

The $x$ 's for which $G (x) = 5 x - 4$ is negative satisfy $x < 4 / 5$ and lie in the interval $(- \infty, 4 / 5)$ .

The $x$ 's satisfying both conditions lie in the intersection of these two intervals, which is empty. No $x$ 's satisfy both these conditions, and no further work needs to be done in this case.

Case 3: $F (x)$ Negative, $G (x)$ Positive

The $x$ 's for which $F (x) = 2 x - 3$ is negative satisfy $x < 3 / 2$ and lie in the interval $(- \infty, 3 / 2)$ .

The $x$ 's for which $G (x) = 5 x - 4$ is positive satisfy $x > 4 / 5$ and lie in the interval $(4 / 5, \infty)$ .

The $x$ 's satisfying both conditions lie in the intersection of these two intervals, and hence lie in $(4 / 5, 3 / 2)$ .

In this interval the inequality

$|2 x - 3| < |5 x - 4|$

becomes

$- (2 x - 3) < 5 x - 4$

$- 2 x + 3 < 5 x - 4$

The solution of this inequality can be obtained by adding $2 x + 4$ to each side, resulting in

$- 2 x + 3 < 5 x - 4$

$2 x + 4 = 2 x + 4$

__________________

$7 < 7 x$

so that $x > 1$ . But not all such $x$ are valid. Only those in the interval $(4 / 5, 3 / 2)$ can be considered. Hence, for the case $F (x) < 0$ and $G (x) > 0$ , the solution of the inequality

$|2 x - 3| < |5 x - 4|$

consists of those $x$ 's in the interval $(1, 3 / 2)$ .

Case 4: $F (x)$ Negative, $G (x)$ Negative

The $x$ 's for which $f (x) = 2 x - 3$ is negative satisfy $x < 3 / 2$ and lie in the interval $(- \infty, 3 / 2)$ .

The $x$ 's for which $g (x) = 5 x - 4$ is negative satisfy $x < 4 / 5$ and lie in the interval $(- \infty, 4 / 5)$ .

The $x$ 's for which both $f (x)$ and $g (x)$ are negative lie in the intersection of these two intervals, and hence lie in $(- \infty, 4 / 5)$ .

If both $f (x)$ and $g (x)$ are negative, then the inequality

$|2 x - 3| < |5 x - 4|$

becomes

$- (2 x - 3) < - (5 x - 4)$

$- 2 x + 3 < - 5 x + 4$

The solution of this inequality can be obtained by adding $5 x$ to each side, and subtracting 3 from each side. This results in

$- 2 x + 3 < - 5 x + 4$

$5 x - 3 = 5 x - 3$

__________________

$3 x < 1$

so that $x < 1 / 3$ . But not all such $x$ are valid. Only those in the interval $(- \infty, 4 / 5)$ can be considered. Hence, for the case when both $f (x)$ and $g (x)$ are negative, the solution of the inequality

$|2 x - 3| < |5 x - 4|$

consists of those $x$ 's in the interval $(- \infty, 1 / 3)$ .

Case 1 gives the interval $(2 / 3, \infty)$ while Case 3 gives the interval $(1, 2 / 3)$ . Since $x = 2 / 3$ satisfies the original inequality (i.e., $0 < |- 2 / 3|$ ), these two cases place the interval $(1, \infty)$ into the solution set. The complete solution of the original inequality is then

$(- \infty, 1 / 3) \cup (1, \infty)$

Figure 1.3.1 highlights the solution set $(- \infty, 1 / 3) \cup (1, \infty)$ with green arrows. These are the values of $x$ for which the graph of $f (x) = |F (x)|$ lies below the graph of $g (x) = |G (x)|$ .

1.3 - Maplet Solution

The inequality

$|2 x - 3| < |5 x - 4|$

can be solved with Inequality Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in Figure 1.3.2.

The left side of the inequality is entered as $f (x)$ , the right, as $g (x)$ . Between these two functions there are radio buttons for selecting the appropriate form of the inequality.

Clicking on the Graph button generates a graph of $f (x)$ (in black), and $g (x)$ (in red). The Plot Options button provides control over the axes on the graph.

Figure 1.3.2 Thumbnail image of Inequality Tutor #1

Clicking on the Intersections button yields $x = 1 / 3, 1$ , the abscissas of the points of intersection.

Clicking on the Solve button generates the solution of the inequality in the form of a union of intervals, namely,

$(- \infty, 1 / 3) \cup (1, \infty)$

To launch Inequality Tutor #1, click the following link: Inequality Tutor #1

1.3 - Interactive Solution

Enter the data for the problem

•	Enter the left-hand side of the inequality. The absolute value template can be found in the Layout palette, or can be entered from the keyboard using the vertical stroke: \|

•	Context Panel: Assign to a Name≻ $f$

•	Enter the right-hand side of the inequality.

•	Context Panel: Assign to a Name≻ $g$

Enter and solve the inequality

•	Write the inequality $f < g$ and press Enter.

•	Context Panel: Solve≻Solve

Calculate the $x$ -coordinate of the intersection

•	Write the equality $f = g$ and press Enter.

•	Context Panel: Solve≻Solve

Obtain Figure 1.3.1

•	Type $f, g$ and press Enter.

•	Context Panel: Plot Builder

Click Global Options button.
Set view for axis[1]: $- 2 \leq x \leq 5$

1.3 - Programmatic Solution

Assign the left-hand side of the inequality to the name $f$

>	$f ≔ \|2 x - 3\|$

Assign the right-hand side of the inequality to the name $g$

>	$g ≔ \|5 x - 4\|$

Solve the inequality.

>	$solve (f < g, \{x\})$

Solve for the intersection of $f (x)$ and $g (x)$

>	$solve (f = g, x)$

Obtain Figure 1.3.1

>	$plot ([f, g], x = - 2 .. 5, color = [black, red])$

Problem 1.4

1.4 - Mathematical Solution

The inequality

$5 \leq |6 x - 1|$

can be solved after the absolute-value function is interpreted according to its definition. Thus, where $6 x - 1$ is positive, $|6 x - 1|$ can be replaced by $6 x - 1$ and the given inequality means

$5 \leq 6 x - 1$

This inequality, valid for $x$ > $1 / 6$ , that is, in the interval $(1 / 6, \infty)$ , can be solved by adding 1 to each side, resulting in $6 \leq 6 x$ , from which follows $x > 1$ . Intersecting the interval $[1, \infty)$ with $(1 / 6, \infty)$ gives the interval $[1, \infty)$ .

Where $6 x - 1$ is negative so that $x < \frac{1}{6}$ , $|6 x - 1|$ must be replaced by $- (6 x - 1)$ so that the given inequality means

Figure 1.4.1 Graph of $f (x) = 5$ (in black) and $g (x) = |6 x - 1|$ (in red)

$5 \leq - (6 x - 1)$

$5 \leq - 6 x + 1$

This inequality can be solved by subtracting 1 from each side, to obtain $4 \leq - 6 x$ . To isolate $x$ , divide by $- 6$ , which will reverse the sense of the inequality to yield $x \leq - 2 / 3$ corresponding to the interval $(- \infty, - 2 / 3]$ .

The solution set for the original inequality is therefore the union of intervals $(- \infty, - 2 / 3] \cup [1, \infty)$ on the real line. Figure 1.4.1 highlights the solution set $(- \infty, - 2 / 3] \cup [1, \infty)$ with green arrows. These are the values of $x$ for which the graph of $f (x)$ lies below the graph of $g (x)$ .

1.4 - Maplet Solution

The inequality

$5 \leq |6 x - 1|$

can be solved with Inequality Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in Figure 1.4.2.

The left side of the inequality is entered as $f (x)$ , the right, as $g (x)$ . Between these two functions there are radio buttons for selecting the appropriate form of the inequality.

Clicking on the Graph button generates a graph of $f (x)$ (in black), and $g (x)$ (in red). The Plot Options button provides control over the axes on the graph.

Figure 1.4.2 Thumbnail image of Inequality Tutor #1

Graphically, the solution of the given inequality consists of those values of $x$ for which the black line (i.e., $f (x)$ ) lies beneath or on the red line (i.e., $g (x)$ ). Clearly, this occurs at, and to the left of the left-hand intersection and at, and to the right of the right-hand intersection.

Clicking on the Intersections button yields $x = - 2 / 3, 1$ , the abscissas of the points of intersection.

Clicking on the Solve button generates the solution of the inequality in the form of a union of intervals, namely,

$(- \infty, - 2 / 3] \cup [1, \infty)$

To launch Inequality Tutor #1, click the following link: Inequality Tutor #1

1.4 - Interactive Solution

Enter the data for the problem

•	Enter the left-hand side of the inequality.

•	Context Panel: Assign to a Name≻ $f$

•	Enter the right-hand side of the inequality. The absolute value template can be found in the Layout palette, or can be entered from the keyboard using the vertical stroke: \|

•	Context Panel: Assign to a Name≻ $g$

Enter and solve the inequality

•	Write the inequality $f \leq g$ and press Enter.

•	Context Panel: Solve≻Solve

Calculate the $x$ -coordinate of the intersection

•	Write the equality $f = g$ and press Enter.

•	Context Panel: Solve≻Solve

Obtain Figure 1.4.1

•	Type $f, g$ and press Enter.

•	Context Panel: Plot Builder

Click Global Options button.
Set view for axis[1]: $- 2 \leq x \leq 2$

•

1.4 - Programmatic Solution

Assign the left-hand side of the inequality to the name $f$

$f ≔ 5$

Assign the right-hand side of the inequality to the name $g$

>	$g ≔ \|6 x - 1\|$

Solve the inequality.

>	$solve (f \leq g, \{x\})$

Solve for the intersection of $f (x)$ and $g (x)$

>	$solve (f = g, x)$

Obtain Figure 1.4.1

>	$plot ([f, g], x = - 2 .. 2, y = - 1 .. 6, color = [black, red])$

Problem 1.5

1.5 - Mathematical Solution

To solve the inequality

$\frac{2}{x} - 1 \leq 7$

begin by adding 1 to each side to obtain

$\frac{2}{x} \leq 8$

or even

$\frac{1}{x} \leq 4$

if both sides of the inequality are then divided by 2. If this were now an equation, the next step would be to multiply both sides by $x$ . However, if $x$ is negative, such a step would reverse the sense of the inequality. Hence, two cases must again be distinguished.

Figure 1.5.1 Graph of $f (x) = 2 / x - 1$ (in black) and $g (x) = 7$ (in red)

	Case 1: $x$ Positive
	If $x$ is positive, then the inequality $\frac{1}{x} \leq 4$ becomes $1 \leq 4 x$ upon multiplication of both sides by $x$ . The solution of this inequality is then $x \geq 1 / 4$ , or the interval $[1 / 4, \infty)$ . The intersection of this interval with the interval $(0, \infty)$ is again the interval $[1 / 4, \infty)$ .

	Case 2: $x Negative$
	If $x$ is negative, then the inequality $\frac{1}{x} \leq 4$ becomes $1 \geq 4 x$ upon multiplication of both sides by $x$ . The solution of this inequality is then $x \leq 1 / 4$ , or the interval $(- \infty, 1 / 4]$ . The intersection of this interval with the interval $(- \infty, 0)$ is the interval $(- \infty, 0)$ .

The complete solution set is then the union of intervals $(- \infty, 0) \cup [1 / 4, \infty)$ on the real line. Figure 1.5.1 highlights the solution set $(- \infty, 0) \cup [1 / 4, \infty)$ with green arrows. These are the values of $x$ for which the graph of $f (x)$ lies below the graph of $g (x)$ .

1.5 - Maplet Solution

The inequality

$\frac{2}{x} - 1 \leq 7$

can be solved with Inequality Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in Figure 1.5.2.

The left side of the inequality is entered as $f (x)$ , the right, as $g (x)$ . Between these two functions there are radio buttons for selecting the appropriate form of the inequality.

Clicking on the Graph button generates a graph of $f (x)$ (in black), and $g (x)$ (in red).

Figure 1.5.2 Thumbnail image of Inequality Tutor #1

The Plot Options button provides control over the axes on the graph.

Graphically, the solution of the given inequality consists of those values of $x$ for which the black line (i.e., $f (x)$ ) lies beneath or on the red line (i.e., $g (x)$ ). Clearly, this occurs to the left of the $y$ -axis, and at, and to the right of the intersection of the curve with the horizontal line.

Clicking on the Intersections button yields $x = 1 / 4$ , the abscissa of the point of intersection.

Clicking on the Solve button generates the solution of the inequality in the form of a union of intervals, namely,

$(- \infty, 0) \cup [1 / 4, \infty)$

(The vertical asymptote at $x = 0$ generates the open interval on the left.)

To launch Inequality Tutor #1, click the following link: Inequality Tutor #1

1.5 - Interactive Solution

Enter the data for the problem

•	Enter the left-hand side of the inequality.

•	Context Panel: Assign to a Name≻ $f$

•	Enter the right-hand side of the inequality. Context Panel: Assign to a Name≻ $g$

Enter and solve the inequality

•	Write the inequality $f \leq g$ and press Enter.

•	Context Panel: Solve≻Solve

Calculate the $x$ -coordinate of the intersection

•	Write the equality $f = g$ and press Enter.

•	Context Panel: Solve≻Solve

Obtain Figure 1.5.1

•	Type $f, g$ and press Enter.

•	Context Panel: Plot Builder

       Click Global Options button.
       Set view for axis[1]: $- 2 \leq x \leq 2$
                            axis[2]: $- 10 \leq y \leq 10$

1.5 - Programmatic Solution

Assign the left-hand side of the inequality to the name $f$

>	$f ≔ \frac{2}{x} - 1$

Assign the right-hand side of the inequality to the name $g$

$g ≔ 7$

Solve the inequality.

>	$solve (f \leq g, \{x\})$

Solve for the intersection of $f (x)$ and $g (x)$

>	$solve (f = g, x)$

Obtain Figure 1.5.1

>	$plot ([f, g], x = - 2 .. 2, y = - 10 .. 10, color = [black, red])$

Problem 1.6

1.6 - Mathematical Solution

The analytic solution of the inequality

$|\frac{2}{x} - 1| \leq 7$

must be split into the two cases

$\frac{2}{x} - 1 < 0$ putting $x$ in $(- \infty, 0) \cup (2, \infty)$

and

$\frac{2}{x} - 1 > 0$ putting $x$ in $(0, 2)$

The inequality $\frac{2}{x} - 1 < 0$ is solved by transforming it to

$\frac{2}{x} < 1$

(add 1 to both sides), then making the dichotomy

Figure 1.6.1 Graph of $f (x) = |2 / x - 1|$ (in black) and $g (x) = 7$ (in red)

$x < 0 \Rightarrow 2 > x$ or $(- \infty, 2) \cap (- \infty, 0) = (- \infty, 0)$

$x > 0 \Rightarrow 2 < x$ or $(2, \infty) \cap (0, \infty) = (2, \infty)$

The inequality $\frac{2}{x} - 1 > 0$ is solved by transforming it to $\frac{2}{x} > 1$ (add 1 to both sides), then making the dichotomy

$x < 0 \Rightarrow 2 < x$ or $(2, \infty) \cap (- \infty, 0) = \emptyset$

$x > 0 \Rightarrow 2 > x$ or $(- \infty, 2) \cap (0, \infty) = (0, 2)$

Case 1: $\frac{2}{x} - 1$ is Negative

If $\frac{2}{x} - 1 < 0$ so that $x$ lies in $(- \infty, 0) \cup (2, \infty)$ , then the original inequality becomes

$- (\frac{2}{x} - 1) \leq 7$

$1 - \frac{2}{x} \leq 7$

which can further be manipulated into the form

$- 6 \leq \frac{2}{x}$

$- 3 \leq \frac{1}{x}$

When $x$ is in $(- \infty, 0)$ , this inequality becomes

$- 3 x$ > 1

so that

$x \leq - 1 / 3$

These calculations add the interval $(- \infty, 0) \cap (- \infty, - 1 / 3) = (- \infty, - 1 / 3)$ to the solution set. However, at $x = - 1 / 3$ the original inequality becomes

$|- 6 - 1| \leq 7$

so that $x = - 1 / 3$ is in the solution set. Therefore, the interval can be written as $(- \infty, - 1 / 3]$ .

When $x$ is in $(2, \infty)$ , the inequality $- 3 \leq \frac{1}{x}$ becomes $- 3 x \leq 1$ so that $x > - 1 / 3$ . The intersection of the intervals $(2, \infty)$ and $(- 1 / 3, \infty)$ is the interval $(2, \infty)$ . However, at $x = 2$ , the original inequality becomes $|1 - 1| \leq 7$ so that $x = 2$ is in the solution set and we write this last interval as $[2, \infty)$ .

	Case 2: $\frac{2}{x} - 1$ is Positive
	If $\frac{2}{x} - 1 > 0$ so that $x$ lies in $(0, 2)$ , then the original inequality becomes $\frac{2}{x} - 1 \leq 7$ or $\frac{2}{x} \leq 8$ or even $\frac{1}{x} \leq 4$ Since $x$ is positive in the interval $(0, 2)$ , this last inequality becomes $1 \leq 4 x$ and then $1 / 4 \leq x$ Thus, the solution set is augmented by the interval $(0, 2) \cap (1 / 4, \infty) = (1 / 4, 2)$ .

The complete solution to the original inequality now contains the intervals $(- \infty, - 1 / 3]$ , $(1 / 4, 2)$ , and $[2, \infty)$ . At $x = 1 / 4$ the original inequality becomes $|8 - 1| \leq 7$ , so the point $x = 1 / 4$ is added to the solution set and the complete solution is the union $(- \infty, - 1 / 3] \cup [1 / 4, \infty)$ on the real line. Figure 1.6.1 highlights the solution set $(- \infty, - 1 / 3] \cup [1 / 4, \infty)$ with green arrows. These are the values of $x$ for which the graph of $f (x)$ lies below the graph of $g (x)$ .

1.6 - Maplet Solution

The inequality

$|\frac{2}{x} - 1| \leq 7$

can be solved with Inequality Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in Figure 1.6.2.

The left side of the inequality is entered as $f (x)$ , the right, as $g (x)$ . Between these two functions there are radio buttons for selecting the appropriate form of the inequality.

Clicking on the Graph button generates a graph of $f (x)$ (in black), and $g (x)$ (in red).

Figure 1.6.2 Thumbnail image of Inequality Tutor #1

The Plot Options button provides control over the axes on the graph.

Graphically, the solution of the given inequality consists of those values of $x$ for which the black line (i.e., $f (x)$ ) lies beneath or on the red line (i.e., $g (x)$ ). Clearly, this occurs at, and to the left of the left-hand intersection of the curve and horizontal line, and at, and to the right of the right-hand intersection of the curve and the horizontal line.

Clicking on the Intersections button yields $x = - \frac{1}{3}, \frac{1}{4}$ , the abscissas of the points of intersection.

Clicking on the Solve button generates the solution of the inequality in the form of a union of intervals, namely,

$(- \infty, - 1 / 3] \cup [1 / 4, \infty)$

To launch Inequality Tutor #1, click the following link: Inequality Tutor #1

1.6 - Interactive Solution

Enter the data for the problem

•	Enter the left-hand side of the inequality. The absolute value template can be found in the Layout palette, or can be entered from the keyboard using the vertical stroke: \|

•	Context Panel: Assign to a Name≻ $f$

•	Enter the right-hand side of the inequality. Context Panel: Assign to a Name≻ $g$

Enter and solve the inequality

•	Write the inequality $f \leq g$ and press Enter.

•	Context Panel: Solve≻Solve

Calculate the $x$ -coordinate of the intersection

•	Write the equality $f = g$ and press Enter.

•	Context Panel: Solve≻Solve

Obtain Figure 1.6.1

•	Type $f, g$ and press Enter.

•	Context Panel: Plot Builder

       Click Global Options button.
       Set view for axis[1]: $- 2 \leq x \leq 2$
                            axis[2]: $- 1 \leq y \leq 10$

1.6 - Programmatic Solution

Assign the left-hand side of the inequality to the name $f$

>	$f ≔ \|\frac{2}{x} - 1\|$

Assign the right-hand side of the inequality to the name $g$

$g ≔ 7$

Solve the inequality.

>	$solve (f \leq g, \{x\})$

Solve for the intersection of $f (x)$ and $g (x)$

>	$solve (f = g, x)$

Obtain Figure 1.6.1

>	$plot ([f, g], x = - 2 .. 2, y = - 1 .. 10, color = [black, red])$

Problem 1.7

1.7 - Mathematical Solution

To solve the inequality

$f (x) = |\frac{2}{x} - 1| < |5 x + 8| = g (x)$

define

$F (x) = \frac{2}{x} - 1$

$G (x) = 5 x + 8$

and consider the following four cases.

	$F (x) = 2 / x - 1$	$G (x) = 5 x + 8$
Case 1	Positive	Positive
Case 2	Positive	Negative
Case 3	Negative	Positive
Case 4	Negative	Negative

Figure 1.7.1 Graph of $|f (x)| = | 2 / x - 1 |$ (in black) and $|g (x)| = | 5 x + 8 |$ (in red)

Case 1: $F (x) > 0$ and $G (x) > 0$

From Problem 1.6, $F (x) > 0$ for $x$ in the interval $(0, 2)$ . Also, $G (x) > 0$ for $x > - 8 / 5$ or equivalently, $x$ in the interval $(- 8 / 5, \infty)$ . The intersection of these two intervals is the interval $(0, 2)$ .

In Case 1, the original inequality can be written as $F (x) < G (x)$ so that we have

$\frac{2}{x} - 1 < 5 x + 8$

$\frac{2}{x} < 5 x + 9$

Since $x$ is positive on the common interval $(0, 2)$ , this last inequality can be multiplied by $x$ without changing the sense of the inequality. Hence, we have

$0 < 5 x^{2} + 9 x - 2 = (5 x - 1) (x + 2)$

Both factors must therefore be of the same sign. If both factors are positive, we then have $x > 1 / 5$ and $x > - 2$ . Both these inequalities hold in the interval $(1 / 5, \infty)$ , but $(0, 2) \cap (1 / 5, \infty) = (1 / 5, 2)$ .

If both factors are negative, we then have $x < 1 / 5$ and $x < - 2$ . Both these inequalities hold in the interval $(- \infty, - 2)$ , but the intersection of this interval with $(0, 2)$ is empty.

	Case 2: $F (x) > 0$ and $G (x) < 0$
	From Problem 1.6, $F (x) > 0$ for $x$ in the interval $(0, 2)$ . Also, $G (x) < 0$ for $x < - 8 / 5$ or equivalently, $x$ in the interval $(- \infty, - 8 / 5)$ . The intersection of these two intervals is empty, so there are no values of $x$ for which Case 2 applies.

Case 3: $F (x) < 0$ and $G (x) > 0$

From Problem 1.6, $F (x) < 0$ if $x$ is in $(- \infty, 0) \cup (2, \infty)$ . Also, $G (x) > 0$ for $x > - 8 / 5$ or equivalently, $x$ in the interval $(- 8 / 5, \infty)$ . The intersection

$((- \infty, 0) \cup (2, \infty))$ $\cap (- 8 / 5, \infty) = (- 8 / 5, 0) \cup (2, \infty)$

determines the common domain for the conditions of Case 3. Under these conditions, the original inequality becomes

$- (\frac{2}{x} - 1) < 5 x + 8$

$- \frac{2}{x} < 5 x + 7$

There are two sub-cases to consider. If $x$ is in the interval $(- 8 / 5, 0)$ , then it is negative, and the last inequality becomes

$- 2 > 5 x^{2} + 7 x$

$0 > 5 x^{2} + 7 x + 2 = (5 x + 2) (x + 1)$

The two factors must be of opposite signs. If $5 x + 2 > 0$ while $x + 1 < 0$ , then $x > - 2 / 5$ and $x < - 1$ have no common intersection and the solution in this case would be empty. If, instead, $5 x + 2 < 0$ while $3 x + 1 > 0$ , then $x < - 2 / 5$ and $x > - 1$ are both satisfied on the interval $(- 1, - 2 / 5)$ whose members form a subset of those in the interval $(- 8 / 5, 0)$ .

On the other hand, if $x$ is in the interval $(2, \infty)$ so that it is positive, the inequality

$- \frac{2}{x} < 5 x + 7$

becomes

$0 < 5 x^{2} + 7 x + 2 = (5 x + 2) (x + 1)$

Now, both factors must be of the same sign. If both factors are negative, we have $x < - 2 / 5$ and $x < - 1$ , true in the interval $(- \infty, - 1)$ . However, the intersection of this interval with $(2, \infty)$ is empty. If both factors are positive, we have $x > - 2 / 5$ and $x > - 1$ , true in the interval $(- 2 / 5, \infty)$ . The intersection of this interval with $(2, \infty)$ is the interval $(2, \infty)$ .

Case 4: $F (x) < 0$ and $G (x) < 0$

From Problem 1.6, $F (x) < 0$ if $x$ is in $(- \infty, 0) \cup (2, \infty)$ . Also, $G (x) < 0$ for $x < - 8 / 5$ or equivalently, $x$ in the interval $(- \infty, - 8 / 5)$ . The common domain for Case 4 is then

$((- \infty, 0) \cup (2, \infty)) \cap (- \infty, - 8 / 5) = (- \infty, - 8 / 5)$

On this common interval, $x$ is negative, so the inequality

$- (\frac{2}{x} - 1) < - (5 x + 8)$

becomes

$\frac{2}{x} > 5 x + 9$

and then

$2 < 5 x^{2} + 9 x$

and finally

$0 < 5 x^{2} + 9 x - 2 = (5 x - 1) (x + 2)$

Once again, both factors must be of the same sign. If both are positive, then $x > 1 / 5$ and $x > - 2$ are true on the interval $(1 / 5, \infty)$ . The intersection of this interval with $(- \infty, - 8 / 5)$ is empty. If instead, both factors are negative, then $x < 1 / 5$ and $x < - 2$ , both of which are true on the interval $(- \infty, - 2)$ . The intersection of this interval with $(- \infty, - 8 / 5)$ is the interval $(- \infty, - 2)$ .

Case 1 yielded the interval $(1 / 5, 2)$ .

Case 3 yielded the intervals $(- 1, - 2 / 5)$ and $(2, \infty)$ .

Case 4 yielded the interval $(- \infty, - 2)$ .

At $x = 2$ , the original inequality becomes $|0| < |18|$ , so $x = 2$ is in the solution set. Therefore, we write the solution as

$(- \infty, - 2) \cup (- 1, - 2 / 5)$ $\cup (1 / 5, \infty)$

Figure 1.7.1 highlights the solution set $(- \infty, - 2) \cup (- 1, - 2 / 5)$ $\cup (1 / 5, \infty)$ with green arrows. These are the values of $x$ for which the graph of $f (x)$ lies below the graph of $g (x)$ .

1.7 - Maplet Solution

The inequality

$|\frac{2}{x} - 1| < |5 x + 8|$

can be solved with Inequality Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in Figure 1.7.2.

The left side of the inequality is entered as $f (x)$ , the right, as $g (x)$ . Between these two functions there are radio buttons for selecting the appropriate form of the inequality.

Clicking on the Graph button generates a graph of $f (x)$ (in black), and $g (x)$ (in red).

Figure 1.7.2 Thumbnail image of Inequality Tutor #1

The Plot Options button provides control over the axes on the graph.

Clicking on the Intersections button yields $x = - 2, - 1, - \frac{2}{5}, \frac{1}{5}$ , the abscissas of the points of intersection.

Clicking on the Solve button generates the solution of the inequality in the form of a union of intervals, namely,

$(- \infty, - 2) \cup (- 1, - 2 / 5) \cup (1 / 5, \infty)$

To launch Inequality Tutor #1, click the following link: Inequality Tutor #1

1.7 - Interactive Solution

Enter the data for the problem

•	Enter the left-hand side of the inequality. The absolute value template can be found in the Layout palette, or can be entered from the keyboard using the vertical stroke: \|

•	Context Panel: Assign to a Name≻f

•	Enter the right-hand side of the inequality. The absolute value template can be found in the Layout palette, or can be entered from the keyboard using the vertical stroke: \| Context Panel: Assign to a Name≻ $g$

Enter and solve the inequality

•	Write the inequality $f < g$ and press Enter.

•	Context Panel: Solve≻Solve

Calculate the $x$ -coordinate of the intersection

•	Write the equality $f = g$ and press Enter.

•	Context Panel: Solve≻Solve

Obtain Figure 1.7.1

•	Type $f, g$ and press Enter.

•	Context Panel: Plot Builder

       Click Global Options button.
       Set view for axis[1]: $- 3 \leq x \leq 1$
                           axis[2]: $- 1 \leq y \leq 10$

1.7 - Programmatic Solution

Assign the left-hand side of the inequality to the name $f$

>	$f ≔ \|\frac{2}{x} - 1\|$

Assign the right-hand side of the inequality to the name $g$

>	$g ≔ \|5 x + 8\|$

Solve the inequality.

>	$solve (f < g, \{x\})$

Solve for the intersection of $f (x)$ and $g (x)$

>	$solve (f = g, x)$

Obtain Figure 1.7.1

>	$plot ([f, g], x = - 3 .. 1, y = - 1 .. 10, color = [black, red])$

Problem 1.8

1.8 - Mathematical Solution

To solve the inequality

$5 \leq |\frac{2}{x} - 1|$

define $G (x) = \frac{2}{x} - 1$ . (Note that this is $F (x)$ in Problem 1.7.) From Problems 1.6 and 1.7,

$G (x) > 0$ for $x$ in $(0, 2)$

$G (x) < 0$ for $x$ in $(- \infty, 0) \cup (2, \infty)$

If $G (x) > 0$ , the given inequality becomes

$5 \leq \frac{2}{x} - 1$

$3 < \frac{1}{x}$

Figure 1.8.1 Graph of $f (x) = 5$ (in black) and $| g (x) | = |2 / x - 1|$ (in red)

Since $x$ is positive in this case, we further have

$3 x < 1$

$x < 1 / 3$

This determines the interval $(- \infty, 1 / 3)$ whose intersection with $(0, 2)$ is the interval $(0, 1 / 3)$ .

If $G (x) < 0$ , the given inequality becomes

$5 \leq - (\frac{2}{x} - 1)$

$5 \leq 1 - \frac{2}{x}$

and even

$2 \leq - \frac{1}{x}$

If $x$ is in the interval $(2, \infty)$ so that it is positive, this last inequality becomes

$2 x \leq - 1$

and $x \leq - 1 / 2$ , contradicting the requirement that $x$ be positive. Hence, there is no solution on the interval $(2, \infty)$ .

If $x$ is in the interval $(- \infty, 0)$ so that it is negative, then the inequality

$2 \leq - \frac{1}{x}$

becomes

$2 x \geq - 1$

so that $x > - 1 / 2$ . The intersection of the interval $[- 1 / 2, \infty)$ with $(- \infty, 0)$ is then just $[- 1 / 2, 0)$ .

Figure 1.8.1 highlights the solution set $[- 1 / 2, 0) \cup (0, 1 / 3]$ with green arrows. These are the values of $x$ for which the graph of $f (x)$ lies below the graph of $g (x)$ .

1.8 - Maplet Solution

The inequality

$5 \leq |\frac{2}{x} - 1|$

can be solved with Inequality Tutor #1 .

Clicking this link will launch the tutor with the solution embedded as shown in Figure 1.8.2.

The left side of the inequality is entered as $f (x)$ , the right, as $g (x)$ . Between these two functions there are radio buttons for selecting the appropriate form of the inequality.

Clicking on the Graph button generates a graph of $f (x)$ (in black), and $g (x)$ (in red).

Figure 1.8.2 Thumbnail image of Inequality Tutor #1

The Plot Options button provides control over the axes on the graph.

Graphically, the solution of the given inequality consists of those values of $x$ for which the black line (i.e., $f (x)$ ) lies beneath or on the red curve (i.e., $g (x)$ ). Except for $x = 0$ , this occurs on and between the intersections of the line and the curve.

Clicking on the Intersections button yields $x = - \frac{1}{2}, \frac{1}{3}$ , the abscissas of the points of intersection.

Clicking on the Solve button generates the solution of the inequality in the form of a union of intervals, namely,

$[- 1 / 2, 0) \cup (0, 1 / 3]$

(There are two intervals because of the vertical asymptote at $x = 0$ .)

To launch Inequality Tutor #1, click the following link: Inequality Tutor #1

1.8 - Interactive Solution

Enter the data for the problem

•	Enter the left-hand side of the inequality. Context Panel: Assign to a Name≻f

•	Enter the right-hand side of the inequality. The absolute value template can be found in the Layout palette, or can be entered from the keyboard using the vertical stroke: \| Context Panel: Assign to a Name≻ $G$

Enter and solve the inequality

•	Write the inequality $f < G$ and press Enter.

•	Context Panel: Solve≻Solve

Calculate the $x$ -coordinate of the intersection

•	Write the equality $f = G$ and press Enter.

•	Context Panel: Solve≻Solve

Obtain Figure 1.8.1

•	Type $f, G$ and press Enter.

•	Context Panel: Plot Builder

       Click Global Options button.
       Set view for axis[1]: $- 2 \leq x \leq 3$
                           axis[2]: $- 1 \leq y \leq 7$

1.8 - Programmatic Solution

Assign the left-hand side of the inequality to the name $f$

$f ≔ 5$

Assign the right-hand side of the inequality to the name $g$

>	$g ≔ \|\frac{2}{x} - 1\|$

Solve the inequality.

>	$solve (f < g, \{x\})$

Solve for the intersection of $f (x)$ and $g (x)$

>	$solve (f = g, x)$

Obtain Figure 1.8.1

>	$plot ([f, g], x = - 2 .. 3, y = - 1 .. 7, color = [black, red])$

Problem 1.9

1.9 - Mathematical Solution

The inequality

$3 < 2 x - 3$ $< 7$

can be solved by first adding 3 to each member, so as to obtain

$6 < 2 x$ $< 10$

from which

$3 < x$ $< 5$

follows by division by 2. Thus $x$ must be in the open interval $(3, 5)$ .

Figure 1.9.1 highlights the solution set $3 < x < 5$ with green arrows. These are the values of $x$ for which the graph of $f (x)$ lies between the lines $m (x) = 3$ and $M (x) = 7$ .

Figure 1.9.1 Graph of $m = 3$ (in green), $M = 7$ (in red), and $f (x) = 2 x - 3$ (in black)

1.9 - Maplet Solution

The inequality

$3 < 2 x - 3$ $< 7$

can be solved with Inequality Tutor #2, as per Figure 1.9.2.

Figure 1.9.2 Solution of the inequality $3 < 2 x - 3 < 7$ by Inequality Tutor #2

The left side of the inequality is entered as $m (x)$ , the middle as $f (x)$ , and the right, as $M (x)$ . Between each entry there are radio buttons for selecting the appropriate form of the inequality.

Clicking on the Graph button generates a graph of $f (x)$ (in black), and the horizontal lines $y = a, y = b$ in green and red, respectively.

The Plot Options button provides control over the axes on the graph.

Graphically, the solution of the given inequality consists of those values of $x$ for which the black line (i.e., $f (x)$ ) lies between the red line (i.e., $M (x)$ ) and the green line (i.e. $m (x)$ ). Clearly, this occurs between the intersections of the black line and the two horizontal lines.

Clicking on the $m (x) = f (x)$ button yields $x = 3$ , the abscissa of the point of intersection of the black and green lines.

Clicking on the $f (x) = M (x)$ button yields $x = 5$ , the abscissa of the point of intersection of the black and red lines.

Clicking on the Solve button generates the solution of the inequality in the form of an interval, namely,

$(3, 5)$

To launch Inequality Tutor #2, click the following link: Inequality Tutor #2

1.9 - Interactive Solution

•	The inequality $3 < 2 x - 3 < 7$ implies the two separate inequalities

$3 < 2 x - 3$ and $2 x - 3 < 7$

•	Form a sequence of these two inequalities.

•	Context Panel: Map Command Onto "Map the command" ≻ solve "Append the argument" ≻ $\{x\}$

Obtain Figure 1.9.1

•	Enter the expression $2 x - 3$

•	Context Panel: Plot Builder

•	Drop the numerals 3 and 7 onto the graph

1.9 - Programmatic Solution

Enter the two separate inequalities implicit in the given inequality, $3 < 2 x - 3 < 7$

>	$q ≔ \{3 < 2 x - 3, 2 x - 3 < 7\}$

Solve the pair of inequalities

>	$solve (q, \{x\})$

Obtain Figure 1.9.1.

>	$plot ([3, 2 x - 3, 7], x = 1 .. 6, color = [green, black, red])$

Problem 1.10

1.10 - Mathematical Solution

To solve the inequality

$3 < |\frac{2}{x} - 1|$ $< 7$

define $F (x) = \frac{2}{x} - 1$ as in Problem 1.7. From Problems 1.6 and 1.7,

$F (x) > 0$ for $x$ in $(0, 2)$

$F (x) < 0$ for $x$ in $(- \infty, 0) \cup (2, \infty)$

If $F (x) > 0$ , the given inequality becomes

$3 < \frac{2}{x} - 1$ $< 7$

$4 < \frac{2}{x}$ $< 8$

Figure 1.10.1 Graph of $m = 3$ (in green), $M = 7$ (in red), and $f (x) = |2 / x - 1|$ (in black)

For positive quantities, reciprocation reverses the sense of the inequality. Hence, we find

$\frac{1}{4} > \frac{x}{2} > \frac{1}{8}$

or, upon multiplying by 2,

$\frac{1}{4} < x < \frac{1}{2}$

If $F (x) < 0$ , the given inequality becomes

$3 < - (\frac{2}{x} - 1) < 7$

$3 < 1 - \frac{2}{x} < 7$

and then

$2 < - \frac{2}{x} < 6$

If $x$ is in the interval $(2, \infty)$ so that it is positive, then this inequality is satisfied by no $x$ since it requires a negative number to be bounded by two positive numbers.

If $x$ is in the interval $(- \infty, 0)$ so that it is negative, then the inequality

$2 < - \frac{2}{x} < 6$

or equivalently,

$1 < - \frac{1}{x} < 3$

can be reciprocated since the middle term is positive. The sense of the inequalities will reverse, giving

$1 > - x > \frac{1}{3}$

Multiplying through by $- 1$ then gives

$- 1 < x < - 1 / 3$

In interval form, the complete solution is then the union $(- 1, - 1 / 3) \cup (1 / 4, 1 / 2)$ .

Figure 1.10.1 highlights the solution set in green. These are the values of $x$ for which the graph of $f (x) = |F (x)|$ lies between the lines $m (x) = 3$ and $M (x) = 7$ .

1.10 - Maplet Solution

The inequality

$3 < |\frac{2}{x} - 1| < 7$

can be solved with Inequality Tutor #2, as per Figure 1.10.2.

Figure 1.10.2 Solution of the inequality $3 < |\frac{2}{x} - 1| < 7$ by Inequality Tutor #2

Clicking on the Graph button generates a graph of $f (x)$ (in black), and the horizontal lines $y = a, y = b$ in green and red, respectively.

The Plot Options button provides control over the axes on the graph. By this means, the vertical axis in Figure 1.10.2 was adjusted to $0 \leq y \leq 10$ .

Graphically, the solution of the given inequality consists of those values of $x$ for which the black line (i.e., $f (x)$ ) lies between the red line (i.e., $M (x)$ ) and the green line (i.e. $m (x)$ ). Except for $x = 0$ , this occurs between the intersections of the black curve and the two horizontal lines.

Clicking on the $m (x) = f (x)$ button yields $x = - 1, \frac{1}{2}$ , the abscissas of the points of intersection of the black curve and green lines.

Clicking on the $f (x) = M (x)$ button yields $x = - \frac{1}{3}, \frac{1}{4}$ , the abscissas of the points of intersection of the black curve and red lines.

Clicking on the Solve button generates the solution of the inequality in the form of a union of intervals, namely,

$(- 1, - 1 / 3) \cup (1 / 4, 1 / 2)$

To launch Inequality Tutor #2, click the following link: Inequality Tutor #2

1.10 - Interactive Solution

•	Enter the "middle" of the inequality. The absolute value template can be found in the Layout palette, or can be entered from the keyboard using the vertical stroke: \|

•	Context Panel: Assign to a Name≻f

•	Form a sequence of the two inequalities $3 < f$ and $f < 7$ implied by the given inequality $3 < f < 7$

•	Context Panel: Map Command Onto "Map the command" ≻ solve "Append the argument" ≻ $\{x\}$

Obtain Figure 1.10.1

•	Type $f$ and press Enter.

•	Context Panel: Plots > 2-D plot

•	Drop the numerals 3 and 7 onto the graph

1.10 - Programmatic Solution

Enter the two separate inequalities implicit in the given inequality, $3 < |\frac{2}{x} - 1| < 7$

>	$q ≔ \{3 < \|\frac{2}{x} - 1\|, \|\frac{2}{x} - 1\| < 7\}$

Solve the pair of inequalities

>	$solve (q, \{x\})$

Obtain Figure 1.9.1.

>	$plot ([3, \|\frac{2}{x} - 1\|, 7], x = - 3 .. 3, y = - 1 .. 8, color = [green, black, red])$

	Exercises - Chapter 1
	Solve each of the following inequalities. 1.1. $2 + 5 x \leq 7$ 1.2. $\|9 x - 5\|$ > 4 1.3. $\|5 + 7 x\| < \|4 x - 9\|$ 1.4. $\|8 x - 9\| \leq 5$ 1.5. $3 + \frac{5}{x} < 8$ 1.6. $\|3 + \frac{5}{x}\| \leq 6$ 1.7. $\|3 + \frac{5}{x}\| < \|2 x - 9\|$ 1.8. $\|3 + \frac{5}{x}\|$ > 1 1.9. 4 < $5 x - 3$ < 11 1.10. $2 \leq \|3 + \frac{5}{x}\|$ < 9

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