Stitz-Zeager_College_Algebra_e-book

Prove the second third and fourth parts of theorem 23

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Unformatted text preview: (3, ∞) with [−4, 6] to obtain [−4, −1) ∪ (3, 6]. Graphically, we see the graph of y = |x − 1| is ‘between’ the horizontal lines y = 2 and y = 5 for x 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 2.4 Inequalities 159 y 8 7 6 5 4 3 2 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 x 4. We need to exercise some special caution when solving |x +1| ≥ x+4 . When variables are both 2 inside and outside of the absolute value, it’s usually best to refer to the definition of absolute value, Definition 2.4, to remove the absolute values and proceed from there. To that end, we have |x + 1| = −(x + 1) if x < −1 and |x + 1| = x + 1 if x ≥ −1. We break the inequality into cases, the first case being when x < −1. For these values of x, our inequality becomes −(x + 1) ≥ x+4 . Solving, we get −2x − 2 ≥ x + 4, so that −3x ≥ 6, which means x ≤ −2. 2 Since all of these solutions fall into t...
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