how solve the inequality $−1\leq \leq1$?

Question

How can we solve this equation graphically ? why we talk about the two hyperbole $\frac{1}{x}$ and $\frac{-1}{x}$ ? I divided this equation on two sub-equations $$ xy\leq1$$ and $$xy\geq-1$$.

For the first sub-equation I have two cases when x is greater or not than zero. Also for the second I have two other equations depending if $x \geq0$ or not.

That said, I no longer know how to solve my system of 4 sub equations

Maybe it becomes clearer if you think geometrically, in terms of the level sets of the function $f(x,y)=xy$? — Hans Lundmark, Sep 06 '21 at 06:23

score 6 · Accepted Answer · answered Sep 06 '21 at 06:25

It can be written compactly as $|xy| \le 1$. The region is symmetrical about $x$-axis and $y$-axis due to the absolute value.

Clearly $x=0$ and $y=0$ are part of the solution.

If $x \ne 0$, then we have $|y| \le \frac1{|x|}$

In the first quadrant, it is below the line $y=\frac1x$. The other cases can be obtained by using symmetry.

In the fourth quadrant, it is above $y=-\frac1x$.

In the second quadrant, it is below $y=-\frac1x$.

In the third quadrant, it is above $y=\frac1x$.

score 2 · Answer 2 · answered Sep 06 '21 at 08:08

2

Dividing by cases, we have that

for $x>0$

$$−1\leq \leq1 \iff -\frac1x \le y \le \frac1x$$

for $x<0$ (direction of inequalities flip)

$$−1\leq \leq1 \iff \frac1x \le y \le -\frac1x$$

and for $x=0 \implies xy=0$.

answered Sep 06 '21 at 08:08

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score 0 · Answer 3 · 2021-09-06T08:21:14.567

0

The domain of the solutions is delimited by the two hyperbolas $xy=-1$ and $xy=1$ (each with two branches).

These partition the plane in five regions, one being a "cross" in the middle. As $(0,0)$ is a solution, the whole cross, including boundaries, is the solution set. (There cannot be solutions on the "other side" of a boundary.)

edited Sep 06 '21 at 08:21

answered Sep 06 '21 at 08:14

how solve the inequality $−1\leq \leq1$?

3 Answers3