The solution set of |x-1|+|x+1|

Question

(C) We analyze the inequality $|x-1| + |x+1| < 2$ by considering different intervals for $x$:
$1$. If $x < -1$,then $|x-1| = -(x-1) = -x+1$ and

Vedclass · Accepted Answer

$\phi$

Vedclass · Answer

(C) We analyze the inequality $|x-1| + |x+1| < 2$ by considering different intervals for $x$:
$1$. If $x < -1$,then $|x-1| = -(x-1) = -x+1$ and $|x+1| = -(x+1) = -x-1$. The inequality becomes $(-x+1) + (-x-1) < 2$,which simplifies to $-2x < 2$,or $x > -1$. This contradicts our assumption $x < -1$,so there is no solution in this interval.
$2$. If $-1 \le x < 1$,then $|x-1| = -(x-1) = -x+1$ and $|x+1| = x+1$. The inequality becomes $(-x+1) + (x+1) < 2$,which simplifies to $2 < 2$. This is a false statement,so there is no solution in this interval.
$3$. If $x \ge 1$,then $|x-1| = x-1$ and $|x+1| = x+1$. The inequality becomes $(x-1) + (x+1) < 2$,which simplifies to $2x < 2$,or $x < 1$. This contradicts our assumption $x \ge 1$,so there is no solution in this interval.
Since no value of $x$ satisfies the inequality,the solution set is the empty set,denoted by $\phi$.

The solution set of $|x-1|+|x+1| < 2$ is

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