Precalculus Notes
2.5 Inequalities
Interval Notation
Let c and d be real numbers with c < d.
[c, d] denotes the set of all real numbers x such that c ≤ x ≤ d .
(c, d) denotes the set of all real numbers x such that c < x < d .
[c, d) denotes the set of all real numbers x such that c ≤ x < d .
(c, d] denotes the set of all real numbers x such that c < x ≤ d .
Ex. 1: Represent the interval on a number line.
a) (0, 8]
b) [-2, 1]
0
4
8
-2
c) ( − ∞,0 ]
-0.5
1
-1
0
1
Ex. 2: Use interval notation to denote the set of all real numbers x that satisfy the given
inequality.
a) 5 ≤ x ≤ 8
b) −3 < x < 14
c) x ≥ −8
[5, 8]
(-3, 14)
[-8, ∞)
Ex. 3: Solve the inequality and express your answer in interval notation.
x −1
2x − 1
a) −2 < 4 − 3x ≤ 6
b) 2x + 7 ( 3x − 2 ) < 2 ( x − 1)
c)
+ 2x ≥
+2
4
3
2x + 7 ( 3x − 2 ) < 2 ( x − 1)
⎛ x −1
⎞ ⎛ 2x −1 ⎞
12 ⎜
+ 2x ⎟ ≥ ⎜
+ 2⎟ 12
−2 < 4 − 3x ≤ 6
2x + 21x − 14 < 2x − 2
⎝ 4
⎠ ⎝ 3
⎠
−6 < −3x ≤ 2
23x − 14 < 2x − 2
3( x −1) + 24x ≥ 4 ( 2x −1) + 24
−6 −3x 2
21x < 12
<
≤
3x − 3 + 24x ≥ 8x − 4 + 24
−3 −3 −3
12
27x − 3 ≥ 8x + 20
2
x<
2>x≥−
21
19x ≥ 23
3
4
23
x<
⎡ 2 ⎞
x≥
−
,
2
7
⎟
⎢⎣ 3 ⎠
19
4⎞
⎛
⎡ 23 ⎞
⎜⎝ −∞, ⎟⎠
7
⎢⎣ 19 , ∞⎟⎠
Ex. 4: Solve the inequality. Find exact solutions when possible, and approximate them
otherwise.
a) x 2 − 4x + 3 ≤ 0
( x − 3)( x − 1) ≤ 0
b) x 4 − x 3 − 12x 2 > 4x + 10
x 4 − x 3 − 12x 2 − 4x − 10 > 0
We know the expression will equal 0
when x = 3 or x =1, but we need to find
when it is less than or equal to 0.
We can’t solve this by hand, so let’s graph
it on the calculator and find the zeros. We
then need to find the region(s) above the xaxis, since we’re looking for the region(s)
where the expression is greater than 0.
Look at the graph:
We can see that the parabola goes below the
x-axis between 1 and 3.
Solution: 1 ≤ x ≤ 3 or [1, 3]
Zeros: x ≈ −2.97, 4.21
Solution: x < −2.97 or x > 4.21
( −∞, − 2.97 )  ( 4.21, ∞ )
x2 + x − 2
<0
c) 2
x − 2x − 3
( x + 2 )( x − 1) < 0
( x − 3)( x + 1)
The numerator (and thus the whole fraction)
equals 0 when x = -2 or 1.
When the denominator equals 0, you should
expect a break in the graph, since the function
is undefined at those points.
We are looking for the region(s) below the x-axis, since the fraction needs to be less than 0.
Solution: −2 < x < −1 or 1 < x < 3
( −2, − 1)  (1, 3)