Here we're given
several absolute value inequalities,
and asked to determine which of the given values
are in the solution set or in the interval.
We'll be answering these questions graphically.
Remember the absolute value of a number
is the number's distance from zero.
So for our first question, which of the given values
are in the solution set for the absolute value of x
less than or equal to zero?
The solution set consists of all the real numbers
that are less than or equal to seven units from zero.
So the graph that's on the number line,
we could start at zero, go right seven units,
we can make a closed point, because seven is
less than or equal to seven units from zero,
and then if we go back to zero, and go left seven units,
negative seven is also less than or equal to
seven units from zero.
And all the numbers between these two
have a distance that's less than or equal to
seven units from zero.
Another way to graph this interval,
instead of using closed points,
would be to use square brackets.
It's the same interval, expressed a different way.
Again, the closed brackets indicate the end points
are in the interval.
Now let's go ahead and plot the given values,
and see if they're in the solution set, or not.
So we have x equals zero, which would be here,
which is in the solution set, or interval.
x equals negative one would be here.
Yes, that's in the interval or solution set.
x equals negative two.
Yes, that's in the solution set.
x equals seven would be here.
Because the end point is included,
this value is in the solution set.
So yes.
And 12 would be over here to the far right.
12 is not in the solution set.
Now because the values stay the same
for the remaining three questions,
we'll leave our points here and just change the interval.
Next we have negative absolute value of x
greater than negative two.
Before we graph this though, we do want to solve
for absolute value of x by multiplying and dividing
both sides by negative one.
But remember when doing this,
we do have to reverse the inequality.
So if we multiply both sides by negative one,
we'd have the absolute value of x is not greater than,
but less than positive two.
Again we multiplied both sets by negative one,
and reversed the inequality symbol.
So we want to determine which
of the given values are in the interval.
The absolute value of x less than two,
which would be all the real numbers that are less
than two units from zero.
So again, starting at zero, we go right two units.
This time the end point is not included in the interval,
because two is not less than two units from zero,
it's exactly two.
Then going back to zero, we can go left two units here.
This endpoint is not on the interval though,
so it's open point, and we graph the interval between.
Now, another way to graph this same interval
would be to use rounded parentheses,
to show the end points are not included.
So we can also graph the interval like this.
Again, normally you wouldn't graph both ways.
Different course materials use one or the other.
Now we already have the values plotted.
x equals zero is in the interval, yes.
x equals negative one, yes.
x equals negative two is not here
because the endpoint is not included, so no.
x equals seven, no.
x equals 12, no.
Next question.
Which of the given values are in the solution set
for absolute value of x greater than seven?
So now we're looking for all the numbers in the number line
whose distance from zero are more than seven units.
So starting at zero we'll go over to the right seven units.
This endpoint is not in the interval
because seven is not greater than seven units from zero,
it's exactly seven units.
So we have an open point here,
and then we graph to the right,
because we're looking for the numbers whose distance
from zero is greater than seven units.
Go back to zero, go left seven units.
This endpoint is not in the interval,
but all the numbers to the left have a distance
that is greater than seven units from zero.
Or we could also express this using
a rounded parenthesis here, graph to the left,
as well as an open parenthesis here
and a graph to the right.
Checking our values, x equals zero
is not in the solution set.
Neither is x equals negative one or negative two.
So we're at x equals seven.
These endpoints are not included in the interval,
so seven is not in the solution set.
Notice how the 12 on the far right is,
so only 12 is in the solution set
to our absolute value inequality.
And this last one's a little tricky.
Which of the given values are in the interval
negative absolute value of negative x
less than or equal to five?
First thing we should do though is solve
for the absolute value, so again we're going to multiply
or divide both sides by negative one,
which requires us to reverse the inequality.
So we'd have the absolute value of negative x
is greater than or equal to negative five.
Now for here, remember, absolute value represents
a number's distance from zero,
which we know can't be negative.
So if the value on the left is always non negative,
a non negative number is always going to be
greater than or equal to negative five.
In fact, a non negative number is always going to be
greater than or equal to any negative number.
And therefore the graph of all the numbers
whose distance from zero is greater than or equal to
negative five, would actually be all real numbers,
or all numbers on the number line.
You can pick any real number, pick the opposite of it,
and determine its distance from zero.
It's always going to be greater than or equal to
negative five units, again because distance
is always non negative.
Which means all these values are in the interval,
or the solution set.
I hope you found this helpful.