$|x-1|+|x|+|x+1| \geq 6$ , find the range of values of x

Question

$|x-1|+|x|+|x+1| \geq 6$ , find the range of values of x

Graphing the function,

So $x \in (-\infty,-2)$and $(2,\infty)$ I do not know how to solve this algebraically. I want an efficient method to find the solution set quickly.

Answer 1

You can always break such question in intervals of x and then solve it easily.

Suppose the problem is this : $|x-1| + |x-2| + |x-3|≥6$ I will solve this now I expect that you will solve your problem on your own by understanding how I did this one.

Case 1: $x≥3$

$x−1+x−2+x−3≥6$

$x≥4$

Case 2: $2≤x<3$

$x−1+x−2−x+3≥6$

$x≥6$

Here NO solution because it is not in the range $[2,3)$

Case 3: $1≤x<2$

$x−1−x+2−x+3≥6$

$x≤−2$

Case 4: $x<1$

$−x+1−x+2−x+3≥6$

$x≤0$

so final solution will be $x∈R−(0,4).$

Now try your question you will reach your answer easily if you understood my solution.

|x−1|+|x−2|+|x−3|≥6 : Graph for this Equation (Pink part shows solution area)

Answer 2

Hint: You can divide the problem into 4 cases: when $x < -1$, when $-1 \leq x < 0$, $0 \leq x < 1$, and $1 < x$ to solve the problem

For example, when $-1 \leq x < 0$, we have $|x-1| = 1-x$, $|x| = -x$ and $|x+1| = x+1$

Answer 3

The hint:

It's obvious that $f(x)=|x-1|+|x|+|x+1|$ is a convex even function.

Thus, the equation $f(x)=6$ has two roots maximum.

But $2$ and $-2$ are roots and we obtain the answer.

Answer 4

The function is symmetrical about $x=0$, we can focus on $x \ge 0$.

We study two cases:

• if $0 \le x \le 1 $: then we have $-x+1+x+x+1 \ge 6$ of which, we find no solution.

• If $x > 1$, then $x-1+x+x+1 \ge 6$.

After which, we get the solution for negative $x$ from symmetry.