Question 192510
I should express the solution to
the inequality:

{{{(x+8)/(x-1)>=0}}} 

in interval notation but I always 
get confused with domain and range.
is it (1,infinity)?
<pre><font size = 4 color = "indigo"><b>
No, domain and range finding techniques
aren't involved here.

Since 0 is on the right side,

{{{(x+8)/(x-1)>=0}}}

We begin by finding all critical values, by
setting the numerator = 0 and solving, then
setting the denominator = 0.

Setting the numerator = 0, x+8=0 gives x=-8,
so -8 is one critical value.

Setting the denominator = 0, x-1=0 gives x=1,
so 1 is the other critical value.

Plot these on a number line.  First make all
the circles open, we may have to close some of
them, but for now just use open circles:
  
-------------------o-----------------------------------o-------------
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4 

Now we need to find out what part of the line to shade.
First we pick any test value in the region to the left of -8,
say -9, and substitute it into the inequality

{{{(x+8)/(x-1)>=0}}}

{{{(-9+8)/(-9-1)>=0}}}
{{{(-1)/(-10)>=0}}}
{{{1/10>=0}}}

That is true, so we shade the number line left of -8

<==================o-----------------------------------o-------------
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4

Next we pick any test value in the between -8 and 1,
say 0, and substitute it into the inequality

{{{(x+8)/(x-1)>=0}}}

{{{(0+8)/(0-1)>=0}}}
{{{(8)/(-1)>=0}}}
{{{-8>=0}}}

That is false, so we do not shade the number line between
-8 and 1

Thirdly we pick any test value in the region to the right of 1,
say 2, and substitute it into the inequality

{{{(x+8)/(x-1)>=0}}}

{{{(2+8)/(2-1)>=0}}}
{{{(10)/(1)>=0}}}
{{{10>=0}}}

That is true, so we shade the number line right of 1

<==================o-----------------------------------o============>
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4

Now we must test the critical values to see if we can
shade them or not.

Test critical value -8 by substituting it

{{{(x+8)/(x-1)>=0}}}

{{{(-8+8)/(-8-1)>=0}}}
{{{(0)/(-9)>=0}}}
{{{0>=0}}}

This is true, so we darken the circle at -8

<==================@-----------------------------------o============>
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4

Test critical value 1 by substituting it

{{{(x+8)/(x-1)>=0}}}

{{{(1+8)/(1-1)>=0}}}
{{{9/0>=0}}}

Division by zero is always undefined. So we must leave
the circle at 1 open, so we still have

<==================@-----------------------------------o============>
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4

Now we want to get the interval notation.  On the far left of the 
shading we have negative infinity, -oo and the shading stops at
-8, so we write the left shaded part as

(-oo,-8]

We use a "(" at -oo because infinity is never included.  We use a ] at 
at -8, because -8 is included and has a darkened circle.


Now we put a "union" symbol "U"

(-oo,-8] U

Then the shaded part on the right going
left to right is from 1 to infinity, or (1,oo)

We use "(" at 1 because 1 is not included, so the
final answer is:

(-oo,-8] U (1,oo)

Edwin</pre>