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Solving Inequalities Algebraically

Solving Inequalities Algebraically. Section P.6 – the last section of the chapter!!!. Solving Absolute Value Inequalities. First, take a look at this graph (visually):. y. y = |x|. a. –a. a. x. |x| > a. |x| < a. |x| > a. Solving Absolute Value Inequalities.

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Solving Inequalities Algebraically

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  1. Solving Inequalities Algebraically Section P.6 – the last section of the chapter!!!

  2. Solving Absolute Value Inequalities First, take a look at this graph (visually): y y = |x| a –a a x |x| > a |x| < a |x| > a

  3. Solving Absolute Value Inequalities Let u be an algebraic expression in x and let a be a real number with a > 0. 1. If |u| < a, then u is in the interval (–a, a). That is, |u| < a if and only if –a < u < a 8 8 2. If |u| > a, then u is in the interval (–,–a) or (a, ). That is, |u| > a if and only if u < –a or u > a

  4. Guided Practice Solve |x – 4| < 8 for x. Write your answer in interval notation. Solution: – 4 < x < 12 (– 4, 12)

  5. Solving Quadratic Inequalities First, replace the inequality sign with an equal sign Next, solve for the variable as with any other quadratic equation Finally, examine the graph to see which values of x satisfy the original inequality (often a visual…)

  6. Guided Practice 2 Solve x – x – 12 > 0 for x. Write your answer in interval notation. Now graph: Solution: For what values of x is the graph above zero? Final Solution:

  7. Guided Practice 2 Solve x + 2x + 2 < 0 for x. Write your answer in interval notation. Solve Graph For what values of x is this function below zero? No Solution!!! Nothing below zero!

  8. 3 2 Solve x + 2x – 1 > 0 graphically. Guided Practice: Solution:

  9. Guided Practice: Solve: Solution:

  10. How about a word problem? Basic Projectile Motion (if an object is propelled, and then only subject to the force of gravity) If an object is launched vertically from a point s feet above the ground with an initial velocity of v ft/sec, then the vertical position s (in feet) of the object t seconds after it is launched is: 0 0 s = –16t + v t + s 2 0 0

  11. How about a word problem? A rugby ball is kicked straight up from a height of 3 feet with an initial velocity of 57 ft/sec. When will the ball’s height above ground be 44 feet? Solve: At t = 1 sec (on the way up) and at t = 2.5625 sec (on the way down)

  12. How about another word problem? A rugby ball is kicked straight up from a height of 3 feet with an initial velocity of 57 ft/sec. When will the ball’s height above ground be at least 44 feet? At t between 1 and 2.5625 sec [ 1 , 2.5625 ]

  13. …and another word problem? A projectile is launched straight up from ground level with an initial velocity of 272 ft/sec. 1. When will the projectile’s height above ground be 960 ft? Solve: At t = 5 sec (on the way up) and at t = 12 sec (on the way down)

  14. Whiteboard Problem: 2 Solve 2x + 3x < 20 for x. Write your answer in interval notation. Solution: [ – 4 , 2.5 ]

  15. Whiteboard Problem: 2 Solve x – 4x + 1 > 0 for x. Write your answer in interval notation. 8 Solution: 8 ( – , 0.268] U [ 3.732 , )

  16. Whiteboard problem: 2 Solve :–3x > –2x – 6 Solution:

  17. Whiteboard problem: 3 2 Solve: x + 2x – 4x + 2 < 0 Solution: Homework: p. 59 1-29 odd Tomorrow = review day then Exam Chapter P (first exam!!)

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