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Section 2.7
Question 48
(1/x) ≥ 1/(x + 3)
The above can easily rewritten as
3/[x(x + 3)] ≥ 0.
Set denominator = 0.
By doing so, we get x = 0 and x = -3.
Plot x = 0 and x = -3 on the real number line.
When x = -4, we get -0.25 ≥ -1. True statement.
When x = -2, we get -0.5 ≥ 1.
When x = 2, we get 2 ≤ 5. True statement.
In interval notation, we get the following:
(-infinity, -3) U (0, infinity)
Question 48
(1/x) ≥ 1/(x + 3)
The above can easily rewritten as
3/[x(x + 3)] ≥ 0.
Set denominator = 0.
By doing so, we get x = 0 and x = -3.
Plot x = 0 and x = -3 on the real number line.
When x = -4, we get -0.25 ≥ -1. True statement.
When x = -2, we get -0.5 ≥ 1.
When x = 2, we get 2 ≤ 5. True statement.
In interval notation, we get the following:
(-infinity, -3) U (0, infinity)