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Rational Inequalities Solving Rational Inequalities.

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Presentation on theme: "Rational Inequalities Solving Rational Inequalities."— Presentation transcript:

1 Rational Inequalities Solving Rational Inequalities

2 7/9/2013 Rational Inequalities 2 General Form f(x) = where p(x) and q(x) are polynomials General Strategy Solve related equations p(x) = 0, q(x) = 0 to find boundary points Check sign of f(x) between pairs of consecutive boundary points Rational Inequalities p(x) q(x) > 0... or ≥ or < or ≤

3 7/9/2013 Rational Inequalities 3 Example: Rational Inequalities x y { x  x ≠ 0 } 1 x2x2 = 0 WHY ? Note that we could NOT start by solving Solution set: Since x 2 > 0 for all x then 1 x2x2 > 0 Note that exists for all x 1 x2x2 ∞ = (–,0) (0, ) ∞ ∩ Solve 1 x2x2 > 0 except x = 0 1 x2x2 = y

4 7/9/2013 Rational Inequalities 4 Example: Rational Inequalities { } Could we solve either inequality by rewriting ? Solution set: Since x 2 > 0 for all x then 1 x2x2 > 0 Solve 1 x2x2 ≤ 0 x y 1 x2x2 = y 1 x2x2 ≤ 0 Hence there is NO x for which Question: What if we clear the fraction by writing : ≤ 1 x2x2 (x 2 ) 0 > 1 x2x2 0 OR ?

5 7/9/2013 Rational Inequalities 5 Example: Solve Note that x ≠ 2 Since (x – 2) 2 ≥ 0 for all x and 2x > 0 provided x ≠ 0 then for all x > 0 except for x = 2 Rational Inequalities x y 2x (x – 2 ) 2 > 0 2x (x – 2 ) 2 > 0 x = 2 Solution set: (0, 2) (2, ) ∞ ∩ = { x  0 < x < 2 } { x  2 < x } ∩

6 7/9/2013 Rational Inequalities 6 Example: Note that x ≠ 2 and x ≠ – 2 Rational Inequalities x y 2 – x 3 > x 2 + x = x g(x) 3 2 – x f(x) =, x = –2 x = 2 f(x) y = 1 g(x) Question: If Can we solve by solving the related equation for boundary points ? Solution set: { x | –2 < x < 2 } = (–2, 2) Clearly f(x) > g(x) only for then for what x is f(x) > g(x) ? –2 < x < 2 Solve

7 7/9/2013 Rational Inequalities 7 x y Example: Rational Inequalities 1 x2x2 for all x except x = 0 5 x and = 2x 2 – 5x + 2 x2x2 y = 2 Note: Solve 5 x ≥ 0 2 x2x2 + 2 f(x) = – exist Assuming x ≠ 0 then ≥ 0 5 x 2 x2x2 + 2 f(x) = – So f(x) has sign of the numerator For x ≠ 0 denominator is positive ●

8 7/9/2013 Rational Inequalities 8 Example: Rational Inequalities ≥ 0 = 2x 2 – 5x + 2 x2x2 f(x) x y y = 2 ● So f(x) has the sign of the numerator Set numerator to zero: 2x 2 – 5x + 2 = 0 Solutions: and … so boundary points at x = 0, x =, x = ● ●

9 7/9/2013 Rational Inequalities 9 x y y = 2 Example: Rational Inequalities ≥ 0 = 2x 2 – 5x + 2 x2x2 f(x) ● x = 0, x =, x = ● ● Checking intervals: x = –1 f(–1) = 9 > 0 x = 0.1 f(.1) = 2.52 > 0 x = 1 f(1) = –1 < 0 x = 3 f(3) = 5 > 0 No Solutions ! Solution set: (–, 0) ∞ [ 2, ) ∞ ∩ ∩ (0, ) 1 2

10 7/9/2013 Rational Inequalities 10 Think about it !


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