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Section 2.7
Question 50
(x^2 + x - 6)/x ≥ 0
Immediately I see that x cannot be zero. To show this fact, we draw a hole on the real number line at x = 0.
Factoring the numerator, I get (x - 2)(x + 3).
Setting to zero, I get x = 2 and x = -3.
See number line below.
When x = -4, we get -3/2 ≥ 0. False statement.
When x = -3, we get 0 ≥ 0. True statement. The hole is filled at x = -3.
When x = -2, we get 2 ≥ 0. True statement.
When x = 1, we get -4 ≥ 0. False statement.
When x = 2, we get 0 ≥ 0. True statement. The hole is filled at x = 2.
When x = 3, we get 2 ≥ 0.
The following intervals satisfy the original inequality:
[-3, 0) U (0, -2] U [2, infinity)
Question 50
(x^2 + x - 6)/x ≥ 0
Immediately I see that x cannot be zero. To show this fact, we draw a hole on the real number line at x = 0.
Factoring the numerator, I get (x - 2)(x + 3).
Setting to zero, I get x = 2 and x = -3.
See number line below.
When x = -4, we get -3/2 ≥ 0. False statement.
When x = -3, we get 0 ≥ 0. True statement. The hole is filled at x = -3.
When x = -2, we get 2 ≥ 0. True statement.
When x = 1, we get -4 ≥ 0. False statement.
When x = 2, we get 0 ≥ 0. True statement. The hole is filled at x = 2.
When x = 3, we get 2 ≥ 0.
The following intervals satisfy the original inequality:
[-3, 0) U (0, -2] U [2, infinity)
