2. To solve a rational inequality, we factor the numerator and the denominator into irreducible factors. The cut points are the real of the numerator and the real. denominator. Then we find the intervals determined by the -, and we use test points to find the sign of the rational function on each interval. Let (x + 2)(x – 1) r(x) = (x – 3)(x + 4) Fill in the diagram below to find the intervals on which r(x) 2 0. 3 Sign of x + 2 x - 1 x - 3 x + 4 (x + 2)(x – 1) (x – 3)(x + 4) From the diagram we see that r(x) 2 0 on the intervals , and
2. To solve a rational inequality, we factor the numerator and the denominator into irreducible factors. The cut points are the real of the numerator and the real. denominator. Then we find the intervals determined by the -, and we use test points to find the sign of the rational function on each interval. Let (x + 2)(x – 1) r(x) = (x – 3)(x + 4) Fill in the diagram below to find the intervals on which r(x) 2 0. 3 Sign of x + 2 x - 1 x - 3 x + 4 (x + 2)(x – 1) (x – 3)(x + 4) From the diagram we see that r(x) 2 0 on the intervals , and
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter3: Polynomial And Rational Functions
Section3.7: Polynomial And Rational Inequalities
Problem 2E
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Question
To solve a rational inequality, we factor the numerator and
the denominator into irreducible factors. The cut points are
the real of the numerator and the real
denominator. Then we find the intervals determined by the
, and we use test points to find the sign
of the rational function on each interval. Let
Transcribed Image Text:2. To solve a rational inequality, we factor the numerator and
the denominator into irreducible factors. The cut points are
the real
of the numerator and the real.
denominator. Then we find the intervals determined by the
-, and we use test points to find the sign
of the rational function on each interval. Let
(x + 2)(x – 1)
r(x) =
(x – 3)(x + 4)
Fill in the diagram below to find the intervals on which
r(x) 2 0.
3
Sign of
x + 2
x - 1
x - 3
x + 4
(x + 2)(x – 1)
(x – 3)(x + 4)
From the diagram we see that r(x) 2 0 on the intervals
, and
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