Rational Inequality...5

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Section 2.7
Question 48

20211003_182855.jpg


(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)

20211003_184515.jpg
 
perfect

here is a graph

MSP8324g79cef3gcd10e500004564edb51a9f8hde

Cool. Section 2.7 does not say to graph rational inequalities on the xy-plane. Ron Larson expects us to graph solution sets on the real number line. I think graphing rational functions on the xy-plane is still a few chapters away but thanks for the graph. In fact from now on, graph the function for me if I get the right answer.
 

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