(x + 1)(x + 2) > 0
We find the critical values of that which are the solutions to
the equation which is found by replacing the > by =:
(x + 1)(x + 2) = 0
x + 1 = 0 x + 2 = 0
x = -1; x = -2
We put those two critical nubers on a number line:
----------o--o-------------------
-5 -4 -3 -2 -1 0 1 2 3 4 5
We test a value for x less that -2, say x = -3
(x + 1)(x + 2) > 0
(-3 + 1)(-3 + 2 > 0
(-2)(-1) > 0
2 > 0
That is true so we shade the number to the left of -2
<=========o--o-------------------
-5 -4 -3 -2 -1 0 1 2 3 4 5
Test a value for x between -2 and -1, say x = -1.5
(x + 1)(x + 2) > 0
(-1.5 + 1)(-1.5 + 2 > 0
(-0.5)(0.5) > 0
-0.25 > 0
That is false so we do not shade the number line between -3 and -2.
We test a value for x greater that -1, say x = 0
(x + 1)(x + 2) > 0
(0 + 1)(0 + 2 > 0
(1)(2) > 0
2 > 0
That is true so we shade the number line to the right of -1
<=========o--o==================>
-5 -4 -3 -2 -1 0 1 2 3 4 5
x < -2 or x > -1 [choice (e)]
The interval notation is an abbreviation for that number line:
(-∞, -2) U (-1, ∞)
Edwin
