College Algebra, 2013a

212 Linear and Quadratic Functions
y
4
3
y =3
2
y = |x − 1|
−4 −3 −2 −1 1 2 3 4 5
x
We see that the graph of y = |x − 1| is above the horizontal line y =3forx4
hence this is where |x − 1| > 3. The two graphs intersect when x = −2 andx =4,sowehave
graphical confirmation of our analytic solution.
2. To solve 4 − 3|2x +1| > −2 analytically, we first isolate the absolute value before applying
Theorem 2.4. Tothatend,weget−3|2x +1| > −6 or|2x +1| < 2. Rewriting, we now have
−2 < 2x +1< 2sothat− 3 2 2sox − 1 < −2 orx − 1 > 2. We get x3. Our solution to the first
inequality is then (−∞, −1)∪(3, ∞). For |x−1| ≤5, we combine results in Theorems 2.1 and
2.4 to get −5 ≤ x − 1 ≤ 5sothat−4 ≤ x ≤ 6, or [−4, 6]. Our solution to 2 < |x − 1| ≤5is
comprised of values of x which satisfy both parts of the inequality, so we take the intersection 1
of (−∞, −1) ∪ (3, ∞) and[−4, 6] to get [−4, −1) ∪ (3, 6]. Graphically, we see that the graph
of y = |x − 1| is ‘between’ the horizontal lines y = 2 and y =5forx values between −4 and
−1 as well as those between 3 and 6. Including the x values where y = |x − 1| and y =5
intersect, we get
1 See Definition 1.2 in Section 1.1.1.