Solving $|x-1|+|2-x|>3+x$

Question

So I've tried separating this problem into three cases

First Case

$$\begin{cases}(x-1)+(2-x)>3+x...(a) \\x\geq2...(b) \end{cases}$$ from (a)

$\Rightarrow x<-2 $

From $(a)\cap(b)$

$x\in \emptyset $

Second case

$$\begin{cases}(x-1)-(2-x)>3+x...(c) \\ x \in [1,2)...(d) \end{cases}$$

From (c)

$\Rightarrow x>6$

From $(c)\cap(d)$

$x \in \emptyset$

Third and final case

$$\begin{cases} -(x-1)-(2-x)>3+x...(e) \\ x<1...(f) \end{cases}$$

From (e)

$\Rightarrow x<-4 $

From $(e)\cap(f)$

$\Rightarrow x<-4$

So my final solution should be the union of all these cases

$\Rightarrow x<-4$

But wolfram says solutions are $x<0 \cup x>6$

I can't seem to find where I made a mistake please help.

Answer 1

You made a mistake by saying $|2-x|=2-x$ when $x\ge2$. The opposite is true.

In your first case, when $x\ge2$ we have $(x-1)+(x-2)>3+x$, which means $x>6$.

In your third case, when $x\lt1$ we have $(1-x)+(2-x)>3+x$, which means $x<0$.

Answer 2

$f(x)=|x-1|+|x-2|-3-x$ is a convex function, which says that the equation $f(x)=0$ has two roots maximum.

But $0$ and $6$ are roots, which gives the answer: $$(-\infty,0)\cup(6,+\infty)$$

Answer 3

Problem is the cases that you have taken. We take the interval for the points where the definition of the modulus function changes, which in this case is at x=1 and x=2. In your case you have considered |f(x)| as + f(x) or -f(x), but whenever you are doing that you need to consider values of x also.

So consider cases x<1 , $1\le x \le 2$ and x>2