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Find $m \in \mathbb R$ for which the equation $|x-1|+|x+1|=mx+1$ has only one unique solution. When does a absolute value equation have only 1 solution?

I solved for $x$ in all 4 cases and got $x=\frac{1}{-m-2},x=\frac{1}{2-m},x=\frac{1}{m},x=-\frac{3}{m}$

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Hint: Draw a graph. – Clive Newstead Nov 25 '12 at 20:20
@CliveNewstead Your comment is good enough to be an answer. – user17762 Nov 25 '12 at 20:31
@CliveNewstead i need to show how i arrived to my answer and i cant use graphs, there must be a way to solve this algebraically, a hint on that maybe? :) – phi Nov 25 '12 at 20:32
up vote 1 down vote accepted

You have $$ |x-1|+|x+1|=\begin{cases} 2x,&\text{ if }x\geq1 \\2,&\text{ if }x\in(-1,1)\\ -2x,&\text{ if }x<-1\end{cases} $$ so $$ g(x)=|x-1|+|x+1|-mx-1=\begin{cases} 2x-mx-1,&\text{ if }x\geq1 \\2-mx-1,&\text{ if }x\in(-1,1)\\ -2x-mx-1,&\text{ if }x<-1\end{cases} $$ For $g(x)=0$ we need:

  1. if $x\geq1$, $(2-m)x-1=0$ requires $1/(2-m)\geq1$ to have a zero, i.e. $1\leq m<2$.
  2. if $x\in(-1,1)$, $2-mx-1=0$, i.e. $mx=1$. This is $x=1/m\in(-1,1)$, so $|m|>1$.
  3. if $x<=1$, $(-2-m)x-1=0$ is $x=1/(-2-m)$; as $x\leq-1$, this is $1/(-2-m)\geq1$, i.e. $-2<m\leq-1$.

In conclusion: if $|m|<1$, there is no possible solution. If $m=1$, there is a single solution (from case 1); if $m=-1$, there is a single solution (from case 3); if $|m|\in(1,2)$, there is a solution from case 2 and another one from either cases 1 and 3.

In short, single solution if and only if $|m|=1$.

(it is a lot easier to understand all this if you draw a picture of $|x-1|+|x+1|$ and of $mx+1$)

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Method 1: Draw a graph to find the answer, and then prove that your answer holds.

Method 2: Solve the equation in the separate cases

  • $x \le -1$ (so that $|x+1|=-(x+1)$ and $|x-1|=-(x-1)$)
  • $-1 \le x \le 1$ (so that etc.)
  • $x \ge 1$ (etc.)

Then play around with the values of $m$ to force only one solution to be valid.

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