Question 289387
<pre><font size = 4 color = "indigo"><b>
{{{abs(1-1/x)<=3}}}

We find the critical values by solving the boundary equation:

{{{abs(1-1/x)=3}}}

Which breaks into these two equations:

{{{1-1/x=3}}}  or {{{1-1/x=-3}}}

{{{x-1=3x}}}   or {{{x-1=-3x}}}

{{{-2x=1}}}    or {{{4x=1}}}

{{{x=-1/2}}}   or {{{x=1/4}}}

We also have a third critical value x=0 because
that is what we get when we set the denominator
in the original problem = 0.  

So we put those points on a number line

----------------o----o--o------------------
-2       -1   -1/2   0 1/4      1         2

We test a value left of -1/2, say -1, in the
original inequality:

{{{abs(1-1/x)<=3}}}
{{{abs(1-1/(-1))<=3}}}
{{{abs(1+1)<=3}}}
{{{abs(2)<=3}}}

That's true, so we shade the region left of -1/2

<================o----o--o------------------
 -2       -1   -1/2   0 1/4      1         2

We test a value between -1/2 and 0, say -.2 in the
original inequality:

{{{abs(1-1/x)<=3}}}
{{{abs(1-1/(-.2))<=3}}}
{{{abs(1+5)<=3}}}
{{{abs(6)<=3}}}
{{{6<=3}}}


That's false, so we DO NOT shade the region between
-1/2 and 0

We test a value between 0 and 1/4, say .2 in the
original inequality:

{{{abs(1-1/x)<=3}}}
{{{abs(1-1/(.2))<=3}}}
{{{abs(1-5)<=3}}}
{{{abs(-4)<=3}}}
{{{4<=3}}}

That's false, so we DO NOT shade the region between
0 and 1/4.

We test a value right of 1/4, say 1, in the
original inequality:

{{{abs(1-1/x)<=3}}}
{{{abs(1-1/(1))<=3}}}
{{{abs(1-1)<=3}}}
{{{abs(0)<=3}}}
{{{0<=3"""

That's true, so we shade the region rightt of 1/4

<================o----o--o=================>
 -2       -1   -1/2   0 1/4      1         2

Now we test the endpoint -1/2

{{{abs(1-1/x)<=3}}}
{{{abs(1-1/(-1/2))<=3}}}
{{{abs(1+2)<=3}}}
{{{abs(3)<=3}}}
{{{3<=3}}}

That's true so we shade -1/2

<================@----o--o=================>
 -2       -1   -1/2   0 1/4      1         2

Now we test the endpoint 1/4

{{{abs(1-1/x)<=3}}}
{{{abs(1-1/(1/4))<=3}}}
{{{abs(1-4)<=3}}}
{{{abs(-3)<=3}}}
{{{3<=3}}}

That's true so we shade 1/4

<================@----o--@=================>
 -2       -1   -1/2   0 1/4      1         2

So the solution set is:

{{{"("}}}{{{-infinity}}}{{{","}}}{{{-1/2}}}{{{"]"}}}{{{U}}}{{{"["}}}{{{1/4}}}{{{","}}}{{{infinity}}}{{{")"}}}

Edwin</pre>