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**Unformatted text preview: **llowing inequalities analytically; check your answers graphically.
1. |x − 1| ≥ 3 3. 2 < |x − 1| ≤ 5 2. 4 − 3|2x + 1| > −2 4. |x + 1| ≥ x+4
2 Solution.
1. To solve |x − 1| ≥ 3, we seek solutions to |x − 1| > 3 as well as solutions to |x − 1| = 3.
From Theorem 2.3, |x − 1| > 3 is equivalent to x − 1 < −3 or x − 1 > 3. From Theorem 2.1,
|x − 1| = 3 is equivalent to x − 1 = −3 or x − 1 = 3. Combining these equations with the
inequalities, we solve x − 1 ≤ −3 or x − 1 ≥ 3. Our answer is x ≤ −2 or x ≥ 4, which, in
interval notation is (−∞, −2] ∪ [4, ∞). Graphically, we have 158 Linear and Quadratic Functions
y
4
3
2 −4 −3 −2 −1 1 2 3 4 5 x We see the graph of y = |x − 1| (the ∨) is above the horizontal line y = 3 for x < −2 and
x > 4, and, hence, this is where |x − 1| > 3. The two graphs intersect when x = −2 and
x = 4, and so we have graphical conﬁrmation of our analytic solution.
2. To solve 4 −...

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