Stitz-Zeager_College_Algebra_e-book

# Prove the second third and fourth parts of theorem 23

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online