Method to solve $|x| + |2-x| \leq x+ 1$

Question

Even if it seems really easy, I'm struggling to solve $$|x| +|2-x |\leq x+1.$$

The book says that $ x \in [1,3] $.

I first rewrote as $x+(2-x)\leq x+1$ with $x\geq 0$ and $-x-(2-x)\leq x+1$ with $x<0$. Then I solved.

For the first, I got $1\leq x$ and for the 2nd, $-3\leq x$

$$x\in ]-3;0[\cup[1;+\infty[ $$

It is maybe a very stupid question, but I can't see what I did wrong? Thanks for your help.

Consider three cases: (i) $x\le0$, (ii) $0<x\le2$, (iii) $x>2$. — George Law, May 17 '17 at 17:47
I will just mention that geometrical meaning of $|x|+|x-2|$ is the sum of distances from $0$ and $2$. So if you move along real line, it should be clear that it is equal to two between $0$ and $2$, So between these two points it is equal to two - it is the length of the line segment from $0$ to $2$. And the graph of $|x|+|x-2|$ is increasing to the right of this interval, decreasing to the left of this interval. You can find a few past question with graphs of similar functions. — Martin Sleziak, Jan 20 '19 at 06:14
One of the answers here has a picture with a similar argument as I mentioned above: How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? Some other posts which have graphs: How to solve this absolute value inequality? $ |x| + |x - 2| \gt 5 $, Solve the following $|x-1| + |x+1|= x-3$ and How could we solve $x$, in $|x+1|-|1-x|=2$? — Martin Sleziak, Jan 20 '19 at 06:18

score 3 · Answer 1 · answered May 17 '17 at 17:52

Note that $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.

Similarly, $|2-x| = 2-x$ if $2 -x \ge 0$ (i.e., if $x \le 2$), and $|2-x| = x-2$ if $2-x < 0$ (i.e., if $x > 2$).

One way is to consider separately the three cases:

$x > 2$ (and so $x \ge 0$ also), so $|x| = x$ and $|2-x| = x-2$.

The inequality becomes $x + x - 2 \le x + 1$.
$0 \le x \le 2$, so $|x| = x$ and $|2-x| = 2-x$.

The inequality becomes $x + 2-x \le x + 1$.
$x < 0$ (and so $x \le 2$ also), so $|x| = -x$ and $|2-x| = 2-x$.

The inequality becomes $-x+2-x \le x + 1$.

Technically there is a fourth combination of $x < 0$ and $x \ge 2$, but clearly this case can't exist since no number can be $ < 0$ and $\ge 2$ at the same time.

Solve the inequalities in each case, and make sure the answer you get makes sense within the restriction of the corresponding case.

For example, for the second case, solving the inequality will give us $x \ge 1$. But remember that this is the case where $0 \le x \le 2$. Thus from the second case we only get $1 \le x \le 2$.

Solve the first and third cases, put it all together, and you'll get $x \in [1,3]$. Let me know if you require further assistance.

score 1 · Answer 2 · answered May 17 '17 at 17:51

1

we can write $$|x|+|x-2|\le x+1$$ the first case: $$x\geq 2$$ then we have to solve $$x+x-2\le x+1$$ second: $$0\le x<2$$ then we have $$x-x+2\le x+1$$ last case: $$x<0$$ then we get $$-x-x+2\le x+1$$ Can you finish?

answered May 17 '17 at 17:51

Dr. Sonnhard Graubner

95,283

Ananth Kamath · Answer 3 · 2019-01-20T05:48:38.560

1

The mistake you have made is that,you have taken $|2-x| = 2-x \forall x\ge 0$, which is not true.

For $x \gt 2$, $|2-x|$ is $ = x-2$.

Hence, you must take intervals as three cases, as George Law points out, $ (i)x\le 0$,

$(ii) 0\lt x\le2$

$(iii) x>2$

edited Jan 20 '19 at 05:48

answered May 17 '17 at 17:56

Ananth Kamath

2,277

Mr. Xcoder · Accepted Answer · 2017-05-17T18:23:32.627

There are 3 cases:

a) $x\leq 0$ (quite useless, because it generates a bright lower bound):

$$|x| +|2-x |\leq x+1\implies -x+ 2-x\leq x+1$$ $$-3x\leq -1$$ $$3x\geq1$$ $$x\geq\frac{1}{3}$$

b) $0<x\leq 2$:

$$|x| +|2-x |\leq x+1\implies x+2-x\leq x+1$$ $$x\geq 1$$

c) $x > 2$:

$$|x| +|2-x |\leq x+1\implies x+x-2\leq x+1$$ $$2x\leq x+3\implies x\leq 3$$

Now, you know the lower bound of $x$, from case b, $x_{min}=1$, and you know the upper bound of $x$ from case c, $x_{max}=3$. Therefore, $1\leq x\leq 3\implies x\in [1;3]$

score 1 · Answer 5 · answered May 17 '17 at 18:45

A change of sign is due to continuity only possible where $$|x| +|2-x |= x+1\iff |2- x|=x+1-|x|.$$ Squaring the equation yields $$2|x|(x-1)=(x-1)(x+3).$$ Since $x=1$ is a solution we may happily square $$2|x|=x+3$$ again leading to $$3(x-3)(x+1)=0.$$ Now consider the truth of $$|x| +|2-x |\leq x+1$$ in the intervals given by the zeroes $-1$, $1$, and $3$.

Method to solve $|x| + |2-x| \leq x+ 1$

5 Answers5