Solve the inequality
and write the solution set in interval notation.
Exclude the points where the expression is undefined. A fraction is only defined when its denominator is non-zero. Here , so
These two points can never belong to the solution set, no matter what the numerator does — a sign never rescues a zero denominator.
Split the absolute value at its breakpoint. changes formula at : it equals when and when . Because the whole inequality behaves differently on the two sides, solve it separately on each and then take the union. Do it algebraically rather than by testing sample numbers, so nothing is missed between test points.
Case 1: . Here , so the numerator collapses to a constant:
For the denominator is positive, so the fraction is negative for every such . The whole ray satisfies the inequality (with itself excluded from step 1).
Case 2: . Here , so the numerator is and
The numerator is zero at and positive for ; the factor is negative throughout this case; the factor changes sign at .
Build the sign table for Case 2. Split the line at the critical values and :
| interval | fraction | |||
|---|---|---|---|---|
Only makes the fraction negative. At the numerator is while the denominator is , so the fraction equals — and is true, so is included.
Take the union of the two cases. Combining from Case 2 with from Case 1:
Spot-checking confirms it: at the value is (true), at it is (false), at it is (true), and at it is (false).
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