Calculus- Early Transcendentals, 2021a

1.1. Algebra 15
• |x| < a means x > −a and x < a (that is, −a < x < a).
• |x|≥a means x ≤−a or x ≥ a.
• |x| > a means x < −a or x > a.
Case 2: a < 0.
• |x| = a has no solutions.
•Both|x|≤a and |x| < a have no solutions.
•Both|x|≥a and |x| > a have solution set {x|x ∈ R}.
Case 3: a = 0.
• |x| = 0hassolutionx = 0.
• |x| < 0 has no solutions.
• |x|≤0hassolutionx = 0.
• |x| > 0 has solution set {x ∈ R|x ≠ 0}.
• |x|≥0 has solution set {x|x ∈ R}.
1.1.6 Solving Inequalities that Contain Absolute Values
We start by solving an equality that contains an absolute value. To do so, we recall that if a ≥ 0thenthe
solution to |x| = a is x = ±a. In cases where we are not sure if the right side is positive or negative, we
must perform a check at the end.
Example 1.16: Absolute Value Equality
Solve for x in |2x + 3| = 2 − x.
Solution. This means that either:
2x + 3 =+(2 − x) or 2x + 3 = −(2 − x)
2x + 3 = 2 − x or 2x + 3 = −2 + x
3x = −1 or x = −5
x = −1/3 or x = −5
Since we do not know if the right side “a = 2 − x” is positive or negative, we must perform a check of our
answers and omit any that are incorrect.
If x = −1/3, then we have LS = |2(−1/3)+3| = |−2/3 + 3| = |7/3| = 7/3 andRS = 2 − (−1/3)=
7/3. In this case LS = RS, sox = −1/3 is a solution.
If x = −5, then we have LS = |2(−5)+3| = |−10 + 3| = |−7| = 7andRS = 2 − (−5)=2 + 5 = 7.
In this case LS = RS,sox = −5 is a solution.
♣
We next look at absolute values and inequalities.