x² ≧ 25
Get 0 on the right side:
x² - 25 ≧ 0
Get the critical numberss by solving the
equation gotten by changing the inequality
symbol to 0.
x² - 25 = 0
(x - 5)(x + 5) = 0
x - 5 = 0; x + 5 = 0
x = 5; x = -5
Place the critical values on a number line:
--------o---------o---------
-5 5
Pick a test value to the left of -5, say -6
Substitute -6 into the inequality
x² - 25 ≧ 0
(-6)² - 25 ≧ 0
36 - 25 ≧ 0
11 ≧ 0
That is true so we shade the region of
the number line to the left of -5
<=======o---------o---------
-5 5
Now pick a test value between -5 and 5, say 0.
Substitute 0 into the inequality
x² - 25 ≧ 0
(0)² - 25 ≧ 0
0 - 25 ≧ 0
-25 ≧ 0
That is false so we DO NOT shade the region
of the number line between -5 and 5. So we
still have:
<=======o---------o---------
-5 5
Pick a test value to the right of 5, say 6
Substitute 6 into the inequality
x² - 25 ≧ 0
(6)² - 25 ≧ 0
36 - 25 ≧ 0
11 ≧ 0
That is true so we shade the region of
the number line to the right of 5
<=======o---------o========>
-5 5
Now we test the critical points themselves:
Testing -5:
x² - 25 ≧ 0
(-5)² - 25 ≧ 0
25 - 25 ≧ 0
0 ≧ 0
That's true, so we darken the circle at -5
<=======⚫--------o========>
-5 5
Testing 5:
x² - 25 ≧ 0
(5)² - 25 ≧ 0
25 - 25 ≧ 0
0 ≧ 0
That's true, so we darken the circle at 5
<=======⚫--------⚫========>
-5 5
The interval notation is an abbreviation
of that graph.
(-∞,-5] U [5,∞)
Edwin
