This is the given inequality I've been trying to solve
$$1/6\leq \frac{1}{\mid x \mid} \leq 1/2$$
However the answer I get is $(0,6] \cup [2,6]$ which is not the answer given in my book. Could you please explain what I did wrong? I split it up into two inequalities $1/6\leq \frac{1}{\mid x \mid}$ and $\frac{1}{\mid x \mid} \leq 1/2$ and solved each individually. I've been trying to find a mistake in my calculations but I don't see one.
$\endgroup$ 24 Answers
$\begingroup$Hint: it is very helpful to flip the fractions (remembering to flip the inequalities): $$\frac{1}{6} \le \frac{1}{|x|} \le \frac{1}{2}\text{ implies } 6 \ge |x| \ge 2.$$
$\endgroup$ 1 $\begingroup$$$\frac{1}{6}\leq\frac{1}{|x|}\leq\frac{1}{2}$$ $$\Leftrightarrow 6\geq|x|\geq 2.$$ Now, $$|x|\leq 6\Leftrightarrow x\in[-6,6]$$ and $$|x|\geq 2\Leftrightarrow x\geq 2$$ or $$x\leq -2.$$ So $x\in[-6,-2]\cup[2,6].$
$\endgroup$ $\begingroup$A completely different approach is to base one's reasoning on a figure like this one:
You can first solve the question for $\frac 16\le \frac 1x\le \frac 12$ and after take the corresponding symmetric in the negative $x$-axis.
Since the function $f(x)=\frac 1x;\space x\gt 0$ is strictly decreasing you have just to solve $\frac 16=\frac 1x$ and $\frac 12=\frac 1x$
Hence $x\in [-6,-2]\cup [2,6]$
$\endgroup$
