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Solving Rational Inequalities

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Presentation on theme: "Solving Rational Inequalities"β€” Presentation transcript:

1 Solving Rational Inequalities
Part 1

2 In order to solve a rational inequality,
it MUST be set equal to ZERO and be simplified to a SINGLE RATIONAL EXPRESSION. Example: 𝒙 𝟐 βˆ’πŸ— 𝒙 𝟐 βˆ’πŸ <𝟎

3 𝒙 𝟐 βˆ’πŸ— 𝒙 𝟐 βˆ’πŸ <𝟎 Now follow these steps… Step 1: Factor the top and bottom of the rational expression and find all the values of x that make the top = 0 and the bottom = 0. (top) (x + 3) (x – 3) so… x = 3 and -3 (bottom) (x + 1) (x – 1) so… x = 1 and -1

4 Step 2: Draw a picture of the x-axis and mark these
points. These are often called β€œcritical points”. Discuss marking the critical points with open or closed circles depending on the inequality symbol.

5 Step 3: Our critical points partition the x-axis into
five intervals. Pick a point (your choice!) within each interval. Let’s use x = 0, x = Β±2, and x = Β±4. Compute f(x) for these points. 𝑓 βˆ’4 = 𝑓 βˆ’2 = 𝑓 0 = 𝑓 2 = 𝑓 4 = βˆ’5 3 <0 7 15 >0 βˆ’9 βˆ’1 >0 βˆ’5 3 <0 7 15 >0

6 These five points represent what happens in the intervals in which they are contained.
You can indicate this on the x-axis by inserting plus or minus signs on the x-axis. + -- + -- +

7 Step 4: We want the expression to be
LESS THAN 0, so we are looking for the intervals that are negative. The solution is …. (-3, -1) and (1, 3). If the inequality had β‰₯ or ≀ then you would use brackets [ ] around the intervals.

8 x – 1 x + 2 ≀ 0 (-2, 1] (-∞, -3] , [-2,-1) , (5, ∞) x 2 + 5x + 6 x 2 βˆ’ 4x – 5 β‰₯ 0 x 2 βˆ’ 16 x 2 βˆ’ 3x + 2 < 0 (-4,1) , (2,4)

9 Solving Rational Inequalities
Part 2 What do I do when the problem isn’t a SINGLE RATIONAL EXPRESSION set equal to ZERO?

10 Solving Rational Inequalities 2
Remember from previous notes that you must COMBINE all fractions together on ONE SIDE so it is set to 0. 2 π‘₯βˆ’4 + 1 π‘₯+1 >0 How would we combine this into a single fraction on the left?

11 Try this one: 1 π‘₯ π‘₯βˆ’4 <0

12 What if there are terms on BOTH sides?
2π‘₯ 4 βˆ’ 5π‘₯+1 3 >3

13 Extra Credit Level:

14 Practice: 1) 2) 3) 4) 5) 4 π‘₯βˆ’6 + 2 π‘₯+1 >0 π‘₯+6 4π‘₯ βˆ’3 β‰₯1 π‘₯ 2 βˆ’π‘₯βˆ’11 π‘₯βˆ’2 ≀3 1 π‘₯ > 1 π‘₯+5 π‘₯ π‘₯+1 βˆ’ π‘₯βˆ’1 π‘₯ < 1 20


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