Presentation is loading. Please wait.

Presentation is loading. Please wait.

7/12/2013Absolute Value Paired Inequalities1 Absolute Value Form for k ≥ 0 | a x + b | > k Absolute Value Inequality a x + b > k –( a x + b) > k Thus a.

Similar presentations


Presentation on theme: "7/12/2013Absolute Value Paired Inequalities1 Absolute Value Form for k ≥ 0 | a x + b | > k Absolute Value Inequality a x + b > k –( a x + b) > k Thus a."— Presentation transcript:

1 7/12/2013Absolute Value Paired Inequalities1 Absolute Value Form for k ≥ 0 | a x + b | > k Absolute Value Inequality a x + b > k –( a x + b) > k Thus a x + b < –k k < a x + b OR A pair of mutually exclusive inequalities 0 k –k a x + b Solution Set for a x + b ) ) This works for k ≥ 0 … what about k < 0 ? a x + b

2 7/12/2013Absolute Value Paired Inequalities2 Absolute Value Form for k < 0 | a x + b | > k Absolute Value Inequality a x + b > k –( a x + b) > k a x + b < –k k < a x + b OR A pair of overlapping inequalities 0 k –k a x + b Solution Set for a x + b ) ) a x + b Question: Is this false for any x ?

3 7/12/2013Absolute Value Paired Inequalities3 Absolute Value Form for all k | a x + b | > k Thus for k ≥ 0 Absolute Value Inequality 0 k –k a x + b Solution Set for a x + b ) ) a x + b 0 k –k a x + b Solution Set for a x + b ) ) a x + b … but where is x ? … and x For k < 0


Download ppt "7/12/2013Absolute Value Paired Inequalities1 Absolute Value Form for k ≥ 0 | a x + b | > k Absolute Value Inequality a x + b > k –( a x + b) > k Thus a."

Similar presentations


Ads by Google