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Can somebody explain to me (or link me a site which does) how to solve this? $$ ||x+1| -1| \geq 3 $$ I have no idea how to work out this double absolute value sign.

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Begin with $|x+1|$, and consider the cases $x+1\geq 0$ (1) and $x+1\leq 0$ (2). In the case $x+1\geq 0$, we have that $|x+1|=x+1$, so $||x+1|-1|=|x|$. Now, solve $|x|\geq 3$, with $x+1\geq 0$ as extra condition. Then, do the same for (2). –  barto Oct 6 '12 at 17:50

4 Answers 4

up vote 3 down vote accepted

Work from the outside in. To start with you have the inequality $|u-1|\ge 3$, where $u=|x+1|$. This is equivalent to $$u-1\le -3\quad\mathbf{or}\quad u-1\ge 3\;.$$ Solving these, we find that $$u\le-2\quad\mathbf{or}\quad u\ge 4\;,$$ which in terms of the original variable $x$ is

$$|x+1|\le -2\quad\mathbf{or}\quad|x+1|\ge 4\;.\tag{1}$$

It’s impossible for an absolute value to be negative, so $(1)$ reduces to $|x+1|\ge 4$. As in the very first step, this is equivalent to

$$x+1\le-4\quad\mathbf{or}\quad x+1\ge 4\;,$$ whose solution is

$$x\le-5\quad\mathbf{or}\quad x\ge 3\;.$$

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Thank you very much! Using WolframAlpha it returns me a different value wolframalpha.com/input/?i=%7C%7Cx+%2B+1%7C+-1%7C+%3E%3D+5. Can you eplain me how to see those result graphically? Why the module graph overturns from (-2, 0)? Thanks –  l_core Oct 6 '12 at 18:02
    
@l_core because you are solving $||x+1|-1|\geq 5$ there instead of $||x+1|-1|\geq 3$ –  no identity Oct 6 '12 at 18:04
    
You are right! ;) –  l_core Oct 6 '12 at 18:09
    
@l_core: The sharp corner at $x=-1$ results from the $|x+1|$ inside the inequality: that function has a sharp corner at $x=-1$. The sharp corners at $x=-2$ and $x=0$ are from the outer absolute value: it has a sharp corner where $|x+1|=1$. Here’s the correct picture. –  Brian M. Scott Oct 6 '12 at 18:11

If I were in front of the class that got this problem, I’d graph it, analyzing it from the inside out. First, $|x+1|$ has a graph that’s vee-shaped, with its vertex at $(-1,0)$. Then $|x+1|-1$ equally has a vee-shaped graph, with its vertex at $(-1,-1)$. Note that this graph crosses the $x$-axis at $(-2,0)$ and the origin. When you take the absolute value of the function $|x+1|-1$, the part of the graph that's below the $x$-axis gets reflected above. The result now is a W-shaped graph, with the left angle at $(-2,0)$, the middle angle at $(-1,1)$, and the right angle at the origin. To the left of $(-2,0)$, the graph rises (as you walk away from the origin) with slope $-1$, i.e. has the equation $y=-x-2$, while to the right of the origin, the graph rises with slope $1$, i.e. it’s just $y=x$. For the value of this function to be at least $3$, you need $x\ge3$ on the right, and $x\le-5$ on the left.

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Here is a solution with squaring technique $$ \begin{align} ||x+1|-1|\geq 3&\Longleftrightarrow (|x+1|-1)^2\geq 3^2\quad\text{ since both sides of inequality are non negative}\\ &\Longleftrightarrow (|x+1|-1)^2-3^2\geq 0\\ &\Longleftrightarrow ((|x+1|-1-3)((|x+1|-1)+3)\geq 0\quad\text{ since}\quad a^2-b^2=(a-b)(a+b)\\ &\Longleftrightarrow (|x+1|-4)(|x+1|+2)\geq 0\\ &\Longleftrightarrow |x+1|-4\geq 0\quad\text{ since }\quad\quad|x+1|+2>0\quad\text{for all }x\\ &\Longleftrightarrow |x+1|\geq 4\\ &\Longleftrightarrow |x+1|^2\geq 4^2\quad\text{ since both sides of inequality are non negative}\\ &\Longleftrightarrow (x+1)^2\geq 4^2\quad\text{ since }\quad |a|^2=a^2\\ &\Longleftrightarrow (x+1)^2 - 4^2\geq 0\\ &\Longleftrightarrow ((x+1)-4)((x+1)+4)\geq 0\quad\text{ since}\quad a^2-b^2=(a-b)(a+b)\\ &\Longleftrightarrow (x-3)(x+5)\geq 0\\ &\Longleftrightarrow x\in(-\infty,-5]\cup[3,+\infty) \end{align} $$

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Since you already have an outside in solution, i'll give you the inside out.

  • First, take the case $x+1\geq 0$. Thus, $||x+1|-1|=|x+1-1|=|x|$.

    $|x|\geq3$ has solution $x\leq-3$ or $x\geq3$. Knowing that $x\geq-1$, the first is impossible.

  • Now, consider $x+1\leq0$. Thus, $||x+1|-1|=|-(x+1)-1|=|-x-2|=|x+2|$. $|x+2|\geq3$ has solution $x\leq-5$ or $x\geq1$. Knowing that $x\leq-1$, the second is impossible.

Now you have $x\leq-5$ or $x\geq3$ as final answer.

This way of solving is not the best one. I also prefer the outside-in, but it's just for showing that there are 2 options.

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