Exercise · ZW-ABSOLUTE-VALUE-REAL-010 · ★★★ Advanced
The solution set of \(|x-1|+|x+1|=2\) is:
Choose one option.
Show answer
- D. \([-1,1]\)
Explanation
For \(-1\le x\le1\): \((1-x)+(x+1)=2\) always. Outside that interval the sum exceeds \(2\).
Review the concept: Absolute value
The absolute value \(|x|\) of a real number \(x\) is its distance from zero on the real line; \(|x-a|\) measures the distance between \(x\) and \(a\).
\[|x|=\begin{cases}x,&x\ge0\\-x,&x<0\end{cases}\]
More practice problems: absolute value
- ZW-ABSOLUTE-VALUE-REAL-001 Compute |-7.5|.
- ZW-ABSOLUTE-VALUE-REAL-002 Compute |3-8|+|-2|.
- ZW-ABSOLUTE-VALUE-REAL-003 Solve |x-2|=5.
- ZW-ABSOLUTE-VALUE-REAL-004 The inequality |x-4|<2 is equivalent to:
- ZW-ABSOLUTE-VALUE-REAL-005 Which equality holds for all reals x,y?
- ZW-ABSOLUTE-VALUE-REAL-006 Solve |2x-1|≤5.
- ZW-ABSOLUTE-VALUE-REAL-007 If |x|=3 and |y|=4, what is the largest possible value of |x+y|?
- ZW-ABSOLUTE-VALUE-REAL-008 If |x|=5 and |y|=2, what is the smallest possible value of |x-y|?