Section 1.7 Inequalities

Linear Inequalities

An inequality is linear if each term is constant or a multiple of the variable.

EXAMPLE: Solve the inequality 3x < 9x+ 4 and sketch the solution set.

Solution: We have

3x < 9x+ 4

3x− 9x < 9x+ 4− 9x

−6x < 4(

−1

6

)

(−6x) >

(

−1

6

)

(4)

x > −2

3

The solution set consists of all numbers greater than −2

3. In other words the solution of the

inequality is the interval(

−2

3,∞

)

.

EXAMPLE: Solve the inequalities 4 ≤ 3x− 2 < 13 and sketch the solution set.

1

EXAMPLE: Solve the inequalities 4 ≤ 3x− 2 < 13 and sketch the solution set.

Solution: We have

4 ≤ 3x− 2 < 13

4 + 2 ≤ 3x− 2 + 2 < 13 + 2

6 ≤ 3x < 15(

1

3

)

(6) ≤

(

1

3

)

(3x) <

(

1

3

)

(15)

2 ≤ x < 5

Therefore, the solution set is [2, 5).

EXAMPLE: Solve the inequalities −4 < 5− 3x ≤ 17 and sketch the solution set.

Solution: We have

−4 < 5− 3x ≤ 17

−4− 5 < 5− 3x− 5 ≤ 17− 5

−9 < −3x ≤ 12(

−1

3

)

(12) ≤

(

−1

3

)

(−3x) <

(

−1

3

)

(−9)

−4 ≤ x < 3

Therefore, the solution set is [−4, 3).

2

Nonlinear Inequalities

EXAMPLE: Solve the inequality x2 ≤ 5x− 6 and sketch the solution set.

Solution: The corresponding equation x2 − 5x+6 = (x− 2)(x− 3) = 0 has the solutions 2 and3. As shown in the Figure below, the numbers 2 and 3 divide the real line into three intervals:(−∞, 2), (2, 3), and (3,∞).

✲• •2 3

On each of these intervals we determine the signs of the factors using test values. We choosea number inside each interval and check the sign of the factors x − 2 and x − 3 at the valueselected. For instance, if we use the test value x = 1 from the interval (−∞, 2) shown in Figureabove, then substitution in the factors x− 2 and x− 3 gives

x− 2 = 1− 2 = −1, x− 3 = 1− 3 = −2

Both factors are negative on this interval, therefore x2 − 5x+ 6 = (x− 2)(x− 3) is positive on(−∞, 2). Similarly, using the test values x = 21

2and x = 4 from the intervals (2, 3) and (3,∞),

respectively, we get:

✲• •2 3+ – +

Thus, the solution of the inequality x2 ≤ 5x− 6 is {x | 2 ≤ x ≤ 3} = [2, 3].

3

EXAMPLE: Solve the inequality x(x− 1)(x+ 2) > 0 and sketch the solution set.

Solution: The corresponding equation x(x − 1)(x + 2) = 0 has the solutions 0, 1, and −2. Asshown in the Figure below, the numbers 0, 1, and −2 divide the real line into four intervals:(−∞,−2), (−2, 0), (0, 1), and (1,∞).

✲• • •-2 0 1

On each of these intervals we determine the signs of the factors using test values. We choosea number inside each interval and check the sign of the factors x, x− 1, and x+2 at the valueselected. For instance, if we use the test value x = −3 from the interval (−∞,−2) shown inFigure above, then substitution in the factors x, x− 1, and x+ 2 gives

x = −3, x− 1 = −3− 1 = −4, x+ 2 = −3 + 2 = −1

All three factors are negative on this interval, therefore x(x−1)(x+2) is negative on (−∞,−2).Similarly, using the test values x = −1, x = 1/2 and x = 2 from the intervals (−2, 0), (0, 1)and (1,∞), respectively, we get:

✲• • •-2 0 1– + – +

Thus, the solution of the inequality x(x− 1)(x+ 2) > 0 is (−2, 0) ∪ (1,∞).

EXAMPLE: Solve the inequality x(x− 1)2(x− 3) < 0.

Solution: The corresponding equation x(x − 1)2(x − 3) = 0 has the solutions 0, 1, and 3. Asshown in the Figure below, the numbers 0, 1, and 3 divide the real line into four intervals:(−∞, 0), (0, 1), (1, 3), and (3,∞).

✲• • •0 1 3

On each of these intervals we determine the signs of the factors using test values. We choose anumber inside each interval and check the sign of the factors x and x− 3 at the value selected.For instance, if we use the test value x = −1 from the interval (−∞, 0) shown in Figure above,then substitution in the factors x and x− 3 gives

x = −1, x− 3 = −1− 3 = −4

Both factors are negative on this interval, (x−1)2 is always nonnegative, therefore x(x−1)2(x−3)is positive on (−∞, 0). Similarly, using the test values x = 1/2, x = 2, and x = 4 from theintervals (0, 1), (1, 3), and (3,∞), respectively, we get:

✲• • •0 1 3+ – – +

Thus, the solution of the inequality x(x− 1)2(x− 3) < 0 is (0, 1) ∪ (1, 3).

REMARK: The solution of the inequality x(x− 1)2(x− 3) ≤ 0 is [0, 3].

4

EXAMPLE: Solve the inequality1 + x

1− x≥ 1 and sketch the solution set.

Solution: We have

1 + x

1− x≥ 1

1 + x

1− x− 1 ≥ 0

1 + x

1− x−

1− x

1− x≥ 0

1 + x− (1− x)

1− x≥ 0

2x

1− x≥ 0

As shown in the Figure below, the numbers 0 (at which the numerator of2x

1− xis 0) and 1 (at

which the denominator of2x

1− xis 0) divide the real line into three intervals: (−∞, 0), (0, 1),

and (1,∞).

✲• •0 1

On each of these intervals we determine the sign of2x

1− xusing test values. We choose a

number inside each interval and check the sign of2x

1− xat the value selected. For instance, if

we use the test value x = −1 from the interval (−∞, 0) shown in Figure above, then substitution

in2x

1− xgives

2(−1)

1− (−1)=

−2

2= −1 < 0

Similarly, using the test values x = 1

2and x = 2 from the intervals (0, 1), and (1,∞), respec-

tively, we get:

✲• •0 1– + –

Thus, the solution of the inequality1 + x

1− x≥ 1 is [0, 1).

EXAMPLE: Solve the inequality2x− 3

x+ 1≤ 1 and sketch the solution set.

5

EXAMPLE: Solve the inequality2x− 3

x+ 1≤ 1 and sketch the solution set.

Solution: We have

2x− 3

x+ 1≤ 1

2x− 3

x+ 1− 1 ≤ 0

2x− 3

x+ 1−

x+ 1

x+ 1≤ 0

2x− 3− (x+ 1)

x+ 1≤ 0

2x− 3− x− 1

x+ 1≤ 0

x− 4

x+ 1≤ 0

As shown in the Figure below, the numbers 4 (at which the numerator ofx− 4

x+ 1is 0) and −1 (at

which the denominator ofx− 4

x+ 1is 0) divide the real line into three intervals: (−∞,−1), (−1, 4),

and (4,∞).

✲• •-1 4

On each of these intervals we determine the sign ofx− 4

x+ 1using test values. We choose a

number inside each interval and check the sign ofx− 4

x+ 1at the value selected. For instance,

if we use the test value x = −2 from the interval (−∞,−1) shown in Figure above, then

substitution inx− 4

x+ 1gives

−2− 4

−2 + 1=

−6

−1= 6 > 0

Similarly, using the test values x = 0 and x = 5 from the intervals (−1, 4), and (4,∞),respectively, we get:

✲• •-1 4+ – +

Thus, the solution of the inequality2x− 3

x+ 1≤ 1 is (−1, 4].

6

Absolute Value Inequalities

EXAMPLE: Solve the inequality |x− 5| < 2 and sketch the solution set.

Solution: The inequality |x− 5| < 2 is equivalent to

−2 < x− 5 < 2

−2 + 5 < x− 5 + 5 < 2 + 5

3 < x < 7

The solution set is the open interval (3, 7).

EXAMPLE: Solve the inequality |x− 5| < −2.

Solution: The inequality |x− 5| < −2 has no solutions, since | · | is always nonnegative. So, thesolution set is the empty set.

EXAMPLE: Solve the inequality |7− 3x| < 1.

7

EXAMPLE: Solve the inequality |7− 3x| < 1.

Solution: The inequality |7− 3x| < 1 is equivalent to

−1 < 7− 3x < 1

−1− 7 < 7− 3x− 7 < 1− 7

−8 < −3x < −6

−6

(

−1

3

)

< −3x

(

−1

3

)

< −8

(

−1

3

)

2 < x <8

3

The solution set is the open interval

(

2,8

3

)

.

EXAMPLE: Solve the inequality |3x+ 2| ≥ 4 and sketch the solution set.

Solution: The inequality |3x+ 2| ≥ 4 is equivalent to

3x+ 2 ≥ 4 or 3x+ 2 ≤ −4

3x ≥ 2 3x ≤ −6

x ≥2

3x ≤ −2

The solution set is{

x | x ≤ −2 or x ≥2

3

}

= (−∞,−2] ∪

[

2

3,∞

)

8