# Very simple rational inequality

Sorry for this rather mundane question, but how would I go about solving the following. I know there is a straightforward method to it.

$-3< 1/x \leq 1$

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Obviously $x\not=0$, or else $1/x$ is not defined. Now, consider the two cases $x>0$ and $x<0$. For $x>0$, the inequality $-3<1/x$ always holds. So all you need to solve is $1/x\le1$. Similarly, for $x<0$, you only need to solve $-3<1/x$. – user1551 Sep 30 '11 at 4:38

Welcome to math.stackexchange! This is a forum for maths questions of all levels, so your level is certainly welcome here.

For this inequality, let's do it one part at a time. First let's find all $x$ such that $\displaystyle\frac{1}{x} \leq 1$. First note, we cetainly can't have $x=0$, so if that comes up as part of a solution later, we must exclude it. It may be introduced as a false solution by the following step: Multiply both sides by $x^2$. We do this, rather than just by $x$, because we can be certain that $x^2$ is positive so the sign of the inequality stays the same. Doing this gives us $x\leq x^2$ or $$x(x-1) \geq 0.$$

The simplest way for these types of inequalities is to think of a rough sketch of what the graph of the polynomial looks like. In this case, $x(x-1)$ is non-negative if $x\geq 1$ and if $x\leq 0$. Another way to think about this one is to realize we need both factors to have the same sign for a non-negative product, and for $x\in (0,1)$ that does not occur.

Now the other direction: $\displaystyle-3 < \frac{1}{x}$. Again, multiply by $x^2$ so we have $-3x^2 < x$ or $$x(1+3x)> 0.$$

Again, thinking about the graph, we see this requires $x\in (-1/3,0)$. For the entire inequality, we then require: $x\in (-1/3,0)$ and either a) $x\geq 1$ or b) $x\leq 0$. So then the total inequality holds when $x\in (-1/3,0) .$

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HINT $\rm\:\ 1/x\: =\: 1\ \iff\ x\: =\: 1$

Also $\rm\:\ \ {-}3\: =\: 1/x\ \iff\ x\: =\: -1/3$

So $\rm\: 1/x-1\:$ and $\rm\:1/x+3\:$ don't change sign on the intervals $\rm\:(-\infty,-1/3),\ (-1/3,1),\ (1,+\infty)$ Now test a point in each interval (and the endpoints) to see which satisfy the original inequalities.

For example, the point $\rm\ x = {-1}\in (-\infty, -1/3)\$ does satisfy $\rm\:-3< 1/x \le 1\:$ so the whole interval $\rm\: x < -1/3\:$ is a solution, but the endpoint $\rm\ x = -1/3\$ is not a solution. Next, the point $\rm\: x = 1/2\$ does not satisfy $\rm\: 1/x\le 1\:$ so that excludes the interval $\rm\:(-1/3,1)\:.\:$ Continue in the same way.

This technique of decomposing into constant-sign regions works much more generally (Tarski).

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For this inequality, I think the simplest and most illuminating approach is to draw the graph $y=1/x$ and check which parts of it falls between the two horizontal lines $y=-3$ and $y=1$. (Wolphram Alpha link.)

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