# Thread: Number Line & Intervals

1. ## Number Line & Intervals

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the intervals on a number line.

|x + pi/2| > 1

The |x + pi/2| > 1 means what lies inside the absolute value bars is more than (x + pi/2) units from the origin.

The point (x + pi/2) lies either to the right of 1 or to the left of -1.

How do I show this on a number line?

Is there more algebra involved before plotting on the number line or am I reading too much into the problem?

2. ## Re: Number Line & Intervals

Originally Posted by mathdad1965
The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the intervals on a number line.

|x + pi/2| > 1

The |x + pi/2| > 1 means what lies inside the absolute value bars is more than (x + pi/2) units from the origin.

The point (x + pi/2) lies either to the right of 1 or to the left of -1.

How do I show this on a number line?

Is there more algebra involved before plotting on the number line or am I reading too much into the problem?
the way I would approach this is to try and determine what the general look of

$|x - a| > b,~b \geq 0$

looks like

you know, or should know, that for any function, $f(x)$,

$f(x-a)$ is just that function shifted right by $a$ units

$|x| > b$ is just the union of the reals greater than $b$ and less than $-b$, i.e. $(-\infty, -b) \cup (b, \infty)$

so the solution to $|x-a| > b$ ends up as this union shifted to the right $a$ units, i.e.

$(-\infty, -b+a) \cup (b+a, \infty)$

Applying this to your problem we have

$a=-\dfrac \pi 2,~b=1$

and the solution is

$x \in \left(\infty, -1 -\dfrac \pi 2\right) \cup \left(1-\dfrac \pi 2, \infty \right)$

3. ## Re: Number Line & Intervals

A bit tricky this problem.

4. ## Re: Number Line & Intervals

If you do not know the definitions of the words and symbols it is tricky. If you do know what this means then it is very easy. First, you would know that |x| is x if x is positive and -x if x is negative. In particular, if x+ pi/2 is positive, then |x+ pi/2| is x+ pi/2 so |x+ pi/2|> 1 becomes x+ pi/2> 1. If x+ pi/2 is negative then |x+ pi/2|= -(x+ pi/2) so |x+ pi/2|> 1 becomes -(x+ pi/2)> 1 and then x+ pi/2< -1.

Mark the point 1- pi/2 (about -.57) on the number line. One part of the solution set is all number beyond that. Now mark -1- pi/2 (about -2.57). The rest of the solution set is all numbers lower than that. The closed interval from -1-pi/2 to 1- pi/2 give everything not in the solution set.

5. ## Re: Number Line & Intervals

Originally Posted by HallsofIvy
If you do not know the definitions of the words and symbols it is tricky. If you do know what this means then it is very easy. First, you would know that |x| is x if x is positive and -x if x is negative. In particular, if x+ pi/2 is positive, then |x+ pi/2| is x+ pi/2 so |x+ pi/2|> 1 becomes x+ pi/2> 1. If x+ pi/2 is negative then |x+ pi/2|= -(x+ pi/2) so |x+ pi/2|> 1 becomes -(x+ pi/2)> 1 and then x+ pi/2< -1.

Mark the point 1- pi/2 (about -.57) on the number line. One part of the solution set is all number beyond that. Now mark -1- pi/2 (about -2.57). The rest of the solution set is all numbers lower than that. The closed interval from -1-pi/2 to 1- pi/2 give everything not in the solution set.
Cool. More later....