Slideshow transcript
Slide 1: Finding absolute value inequality statements Copyright ©2007-2008 Home Tuition Services
Slide 2: We can show that if a≥0, −a ≤ x ≤ a ⇔ ∣ x ∣ ≤ a x ≤ −a or x ≥ a ⇔ ∣ x ∣ ≥ a .
Slide 3: Example 1 Find the absolute value inequality statement corresponding to x ≥ −3 or x ≤ 3 . ⇒ x ≥ −3 or x ≤ 3 ⇒ −3 ≤ x ≤ 3 We write in standard≤ form. ⇒ ∣ x ∣≤ 3 The statement is ∣ x ∣ ≤ 3
Slide 4: Example 2 Find the absolute value inequality statement corresponding to x ≤ 3 or x ≥ −3 . This is the same as Example 1 with the inequality statements in reverse order. Don't confuse this with statements x ≤ −3 or x ≥ 3 that lead to a standard ≥ absolute value inequality statement.
Slide 5: ⇒ x ≤ 3 or x ≥ −3 ⇒ −3 ≤ x ≤ 3 We write in standard ≤ form. ⇒ ∣ x ∣≤ 3 The statement is ∣ x ∣ ≤ 3
Slide 6: Example 3 Find the absolute value inequality statement corresponding to x −5 or x 5 . ⇒ x −5 or x 5 Standard form. ⇒ ∣ x ∣ 5 The statement is ∣ x ∣ 5
Slide 7: Example 4 Find the absolute value inequality statement corresponding to −1 x 6 .
Slide 8: ⇒ −1 x 6 5 5 5 ⇒ −1− x− 6− 2 2 2 7 5 7 ⇒ − x− 2 2 2 ∣ ∣ 5 We note the inequality is in a form that 7 ⇒ x− 2 we2can convert to standard form.
Slide 9: ⇒ −1 x 6 Average −1 and 6 . 5 5 5 ⇒ −1− x− 6− −16 5 = 2 2 2 2 2 7 5 7 ⇒ − x− 2 2 2 ∣ ∣ 5 7 ⇒ x− 2 2 5 Subtract from all three sides. 2
Slide 10: ⇒ −1 x 6 5 5 5 ⇒ −1− x− 6− 2 2 2 7 5 7 ⇒ − x− 2 2 2 Standard form. ∣ ∣ 5 7 ⇒ x− 2 2 ∣ ∣ 5 7 The statement is x− 2 2
Slide 11: Example 5 Find the absolute value inequality statement corresponding to −5 x 3 .
Slide 12: ⇒ −5 x 3 ⇒ −5−−1 x−−1 3−−1 ⇒ −4 x1 4 ⇒ ∣ x1 ∣ 4 We note the inequality is in a form that we can convert to standard form.
Slide 13: ⇒ −5 x 3 Average −5 and 3 . ⇒ −5−−1 x−−1 3−−1 −53 = −1 ⇒ −4 x1 4 2 ⇒ ∣ x1 ∣ 4 Subtract −1 from all three sides.
Slide 14: ⇒ −5 x 3 ⇒ −5−−1 x−−1 3−−1 ⇒ −4 x1 4 Standard form. ⇒ ∣ x1 ∣ 4 The statement is ∣ x1 ∣ 4
Slide 15: Example 6 Find the absolute value inequality statement corresponding to x 3 or x 5 .
Slide 16: ⇒ x 3 or x 5 ⇒ x−4 3−4 or x−4 5−4 ⇒ x−4 −1 or x−4 1 ⇒ ∣ x−4 ∣ 1 We note the inequality is in a form that we can convert to standard form.
Slide 17: ⇒ x 3 or x 5 Average 3 and 5 . ⇒ x−4 3−4 or x−4 5−4 35 =4 ⇒ x−4 −1 or x−4 1 2 ⇒ ∣ x−4 ∣ 1 Subtract 4 from all sides.
Slide 18: ⇒ x 3 or x 5 ⇒ x−4 3−4 or x−4 5−4 ⇒ x−4 −1 or x−4 1 Standard form. ⇒ ∣ x−4 ∣ 1 The statement is ∣ x−4 ∣ 1
Slide 19: Example 7 Find the absolute value inequality statement with integer coefficients corresponding to −1 x 6 .
Slide 20: ⇒ −1 x 6 5 5 5 ⇒ −1− x− 6− 2 2 2 7 5 7 ⇒ − x− 2 2 2 ∣ ∣ 5 We note the inequality is in a form that 7 ⇒ x− 2 we2can convert to standard form. ∣ ∣ 5 7 ⇒ 2× x− 2× 2 2 ∣ ∣5 ⇒ 2 x− 7 2 ⇒ ∣ 2 x−5 ∣ 7
Slide 21: ⇒ −1 x 6 Average −1 and 6 . 5 5 5 ⇒ −1− x− 6− −16 5 = 2 2 2 2 2 7 5 7 ⇒ − x− 2 2 2 ∣ ∣ 5 7 ⇒ x− 2 2 5 Subtract 7 from all three sides. ∣ ∣ 5 2× x− 2×2 ⇒ 2 2 ∣ ∣5 ⇒ 2 x− 7 2 ⇒ ∣ 2 x−5 ∣ 7
Slide 22: ⇒ −1 x 6 5 5 5 ⇒ −1− x− 6− 2 2 2 7 5 7 ⇒ − x− 2 2 2 Standard form. ∣ ∣ 5 7 ⇒ x− 2 2 ∣ ∣ Multiply both sides by 5 7 ⇒ 2× x− 2× 2 , the Lowest Common 2 2 ∣ ∣5 Denominator. ⇒ 2 x− 7 2 ⇒ ∣ 2 x−5 ∣ 7
Slide 23: ⇒ −1 x 6 5 5 5 ⇒ −1− x− 6− 2 2 2 7 5 7 ⇒ − x− 2 2 2 ∣ ∣ 5 7 ⇒ x− 2 2 ∣ ∣ Note that a×∣ b ∣=∣ ab ∣ 5 7 ⇒ 2× x− 2× 2 2 if a ≥ 0 . ∣ ∣5 ⇒ 2 x− 7 2 ⇒ ∣ 2 x−5 ∣ 7 The statement is ∣ 2 x−5 ∣ 7
Slide 24: Example 8 Find the absolute value inequality statement corresponding to x2 5 or x−1 4 .
Slide 25: ⇒ x2 5 or x−1 4 ⇒ x 3 or x 5 ⇒ x−4 3−4 or x−4 5−4 ⇒ x−4 −1 or x−4 1 Take care. Isolate x before continuing. ⇒ ∣ x−4 ∣ 1
Slide 26: ⇒ x2 5 or x−1 4 ⇒ x 3 or x 5 ⇒ x−4 3−4 or x−4 5−4 ⇒ x−4 −1 or x−4 1 ⇒ ∣ x−4 ∣ 1 We note the inequality is in a form that we can convert to standard form.
Slide 27: ⇒ x2 5 or x−1 4 ⇒ x 3 or x 5 Average 3 and 5 . ⇒ x−4 3−4 or x−4 5−4 35 =4 ⇒ x−4 −1 or x−4 1 2 ⇒ ∣ x−4 ∣ 1 Subtract 4 from all sides.
Slide 28: ⇒ x2 5 or x−1 4 ⇒ x 3 or x 5 ⇒ x−4 3−4 or x−4 5−4 ⇒ x−4 −1 or x−4 1 Standard form. ⇒ ∣ x−4 ∣ 1 The statement is ∣ x−4 ∣ 1
Slide 29: Example 9 Find the absolute value inequality statement corresponding to 2 x−3 1 or x−2 4 .
Slide 30: ⇒ 2 x−3 1 or x−2 4 ⇒ 2 x 4 or x 6 ⇒ x 2 or x 6 ⇒ 2x 6 ⇒ 2−4 Isolate 6−4 continuing. x−4 x before Take care. ⇒ −2 x−4 2 ⇒ ∣ x−4 ∣ 2
Slide 31: ⇒ 2 x−3 1 or x−2 4 ⇒ 2 x 4 or x 6 ⇒ x 2 or x 6 ⇒ 2x 6 We note we can write in this form ⇒ 2−4 x−4 6−4 in preparation for conversion to standard form. ⇒ −2 x−4 2 ⇒ ∣ x−4 ∣ 2
Slide 32: ⇒ 2 x−3 1 or x−2 4 ⇒ 2 x 4 or x 6 ⇒ x 2 or x 6 ⇒ 2x 6 Average 2 and 6 . ⇒ 2−4 x−4 6−4 26 =4 ⇒ −2 x−4 2 2 ⇒ ∣ x−4 ∣ 2 Subtract 4 from all three sides.
Slide 33: ⇒ 2 x−3 1 or x−2 4 ⇒ 2 x 4 or x 6 ⇒ x 2 or x 6 ⇒ 2x 6 ⇒ 2−4 x−4 6−4 ⇒ −2 x−4 2 Standard form. ⇒ ∣ x−4 ∣ 2 The statement is ∣ x−4 ∣ 2
Slide 34: Finding absolute value inequality statements Copyright ©2007-2008 Home Tuition Services



Add a comment on Slide 1
If you have a SlideShare account, login to comment; else you can comment as a guest- Favorites & Groups
Showing 1-50 of 0 (more)