# More precise way of solving inequality

I need to solve this function: $$\lvert x^2-1\rvert\ge 2x-2\\$$

I solved this equation:

For $x<0$, the solution is non existing, here I got negative root, when I tried to solve quadratic function and for $x\ge 0$ I got points $x_1=-1$ and $x_2=3$.

My question is:

How do I set the solution of equation. Is there any procedure, with wich I can determine is equation valid for $[-1,3]$ or $[-\infty, -1]\lor [3, +\infty]$.

I know that I can just set the numbers and see the result, but I just want to now is there any other different way to do this.

Thanks.

• That's an inequality, not a function. – DonAntonio Dec 30 '13 at 16:31
• Thanks, I wrote the title very fast, so I got a little bit confuesd – depecheSoul Dec 30 '13 at 16:36

As $\displaystyle |x|=\begin{cases} x &\mbox{if } x\ge0 \\-x & \mbox{if } x<0 \end{cases}$

If $x^2-1\ge0\iff x\ge1$ or $x\le-1,$

we get $$x^2-1\ge2x-2\iff x^2-2x+1\ge0\iff (x-1)^2\ge0$$ which is true

If $x^2<1\iff -1<x<1\ \ \ \ (1),$

we get $$-(x^2-1)>2x-2\iff x^2+2x-3<0$$

$$\iff (x+3)(x-1)<0\iff -3<x<1\ \ \ \ (2)$$

Now using $(1),(2)-1<x<1$

So, $x$ can assume any real value

• Another way: As $\displaystyle|x^2-1|\ge0, |x^2-1|\ge 2x-2$ if $2x-2\le0\iff x\le1$ If $\displaystyle x>1, x^2>1\implies |x^2-1|=x^2-1$ we can solve as in the answer – lab bhattacharjee Dec 30 '13 at 16:41
• I dont understand this part of explenation, can you please clarify it. "But as −1<x<1,−1<x<1". Thanks. – depecheSoul Dec 30 '13 at 16:51
• @depecheSoul, edited the answer, sorry for the confusion – lab bhattacharjee Dec 30 '13 at 16:53

Before trying an “automated” method, it's better to have a close look at the problem.

You can first of all observe that if $2x-2<0$, then the inequality obviously holds, because $|x^2-1|\ge0$. So all values $x<1$ will satisfy the inequality.

Now suppose $x\ge 1$. In this case also $x^2\ge1$, so we can write the inequality as $$x^2-1\ge 2x-2$$ which becomes $$x^2-2x+1\ge0$$ or $$(x-1)^2\ge0$$ which is surely true for all $x$, in particular for $x\ge 1$.

Conclusion: the inequality holds for all real $x$.

The "automated" method has been treated in other answers; to summarize it, you can divide the inequality into two and you'll put together the solution sets at the end $$\begin{cases} x^2-1\ge0\\ x^2-1\ge 2x-2 \end{cases} \qquad\text{or}\qquad \begin{cases} x^2-1<0\\ 1-x^2\ge 2x-2 \end{cases}$$ Simple algebraic manipulation transforms them into $$\begin{cases} x\le -1 \text{ or }x\ge1\\ (x-1)^2\ge0 \end{cases} \qquad\text{or}\qquad \begin{cases} -1<x<1\\ x^2+2x-3\le0 \end{cases}$$ and, going on, $$x\le -1\text{ or }x\ge 1 \qquad\text{or}\qquad \begin{cases} -1<x<1\\ -3\le x\le 1 \end{cases}$$ whence the same conclusion as before. (Note: the brace is another way of saying “and”.)

• One question: you are connecting functions with logical operator "OR", so when for example $x^2−1≥2x−2$ is true, the and $1−x^2≥2x−2$ is false, then the function $x^2−1≥2x−2$ is valid. Thanks – depecheSoul Dec 30 '13 at 18:25
• @depecheSoul No, I don't mean that. What's true is that, for example, the original inequation is equivalent to the following “or“ assertion. This means that any solution for one system is a solution of the original inequality (and conversely). – egreg Dec 30 '13 at 18:28

You have two cases :

Case 1 : When $x^2-1\ge0$, you have $x^2-1\ge 2x-2.$

Case 2 : When $x^2-1\lt0$, you have $-(x^2-1)\ge 2x-2.$

The answer is $$"x^2-1\ge0\ \text{and}\ x^2-1\ge 2x-2"\ \text{or}\ "x^2-1\lt0\ \text{and}\ -(x^2-1)\ge 2x-2."$$

Hints:

\begin{align*}|x|<1\implies x^2-1<0\implies |x^2-1|=1-x^2\\ |x|\ge 1\implies x^2-1\ge 0\implies |x^2-1|=x^2-1\end{align*}

So, for example, for $\;x=-1\;$ you get

$$|(-1)^2-1|\ge 2(-1)-2\iff2\ge-4\;\color{green}{\checkmark}$$

• I guess you picked up the \checkmark from my question? :p – user93957 Dec 30 '13 at 16:55
• I honestly have no idea what you're talking about, @Adobe ... – DonAntonio Dec 30 '13 at 17:19

Clearly $x=1$ is a solution. Dividing by $|x-1|$ yields to $$|x+1|\leq2\frac{x-1}{|x-1|},$$ which is obviously true both for $x<1$ and $x>1$.