Inequalities Part 4 Quiz

4 − |x − 2| ≥ 0
or, |x − 2| ≤ 4
or, −4 ≤ x − 2 ≤ 4
or, −2 ≤ x ≤ 6

(x² − 1)/(x − 1) ≥ 0
or, (x − 1)(x + 1) / (x − 1) ≥ 0
For x ≠ 1, x + 1 ≥ 0; or; x ≥ −1
But exclude x = 1 (undefined)
x ≥ −1 and x ≠ 1

(x + 1)² ≤ 9
Take square roots (both sides non-negative, inequality direction preserved):
|x + 1| ≤ 3
or, −3 ≤ x + 1 ≤ 3
or, −4 ≤ x ≤ 2

|2x| > |x + 3|
Square both sides (valid since both are non-negative):
(2x)² > (x + 3)²
or, x² − 2x − 3 > 0
or, (x − 3)(x + 1) > 0
Critical points: x = 3, x = −1
x < −1 or x > 3

(3x − 2)/(x + 4) ≥ 1
or, (3x − 2)/(x + 4) − 1 ≥ 0
or, 2(x − 3)/(x + 4) ≥ 0
or, (x − 3)/(x + 4) ≥ 0
Critical points: x = 3, x = −4 (undefined)
Solution: x < −4 or x ≥ 3

2 − x² ≥ x
or, −x² − x + 2 ≥ 0
or, x² + x − 2 ≤ 0
or, (x + 2)(x − 1) ≤ 0
or, Critical points: x = −2, x = 1
Solution: −2 ≤ x ≤ 1

1/(x + 2) < −1 or, 1/(x + 2) + 1 < 0 or, (x + 3)/(x + 2) < 0 Critical points: x = −3 (numerator = 0), x = −2 (denominator = 0, undefined) −3 < x < −2

|x² − 4| ≤ 5
or, −5 ≤ x² − 4 ≤ 5
or, −1 ≤ x² ≤ 9
Since x² ≥ 0 always, the left part (−1 ≤ x²) is always true.
So we only need: x² ≤ 9
or, −3 ≤ x ≤ 3

x² − 5x + 6 < 0 or, (x − 2)(x − 3) < 0 Critical points: x = 2, x = 3 2 < x < 3

|x − 3| ≥ |x + 1|
Square both sides (valid since both sides are ≥ 0):
or, (x − 3)² ≥ (x + 1)²
or, x² − 6x + 9 ≥ x² + 2x + 1
or, −8x ≥ −8
or, x ≤ 1