Question

Solve the following inequality. Write the answer in interval notation.
Note: If the answer includes more than one interval write the intervals separated by the "union" symbol, U. If needed enter \infty as infinity and -\infty as -infinity . If there are no solutions then type the word "none" without the quotes.
|x + 5| < 6

How do I solve this?

Answer 1

For absolute value, its shows how far a interger is from zero. So you need to set up two inequality. For this one, it would be x+5<6, x-5<6. Then you just solve like a regular inequality.

Answer 2

Kinda sorta, but not quite. Absolute value is distance from zero. So if:

|{something}| < 6

it is equivalent to saying that the {something} is LESS THAN 6 units from 0.

How does something get to be LESS THAN 6 units from 0? It has to be within (but less than) 6 units to the right, or to the left, of 0. Hence:

|{something}| < 6 means that -6< {something} < 6

That is, the goal IS indeed to write compound inequality without abs. value. However, the {something} INSIDE the abs value DOES NOT CHANGE when you set up the compound inequality!!! So the 2 inequalities given above are not quite right, because instead of "sandwiching" the {something} between -6 and 6, it just changed the sign INSIDE the {something} (big no-no, common error), and did not account at all for the part of the inequality what "captures" the lower limit of -6.

The other thing with abs value inequalities is to make sure you correctly write the equivalent compound inequality either as an "and" inequality (which is automatic if you use "between" notation, as I did above) or as an "or" inequality (NOT the case here, but WOULD be the case if you have |{something}|>6). So you can't just make the two inequalities and then " solve like a regular inequality"... you have to solve like the appropriate type of COMPOUND inequality.

Answer 3

So if I had, say:

|x -1| < 9

Then that means that x - 1 is WITHIN 9 units of 0. Hence:

-9 < x - 1 < 9

or if I prefer 2 separate statements joined by "and", instead of "between notation", I can write that as:

-9 < x - 1 AND x - 1 < 9

Notice that I didn't change the {something} inside the abs value signs at all. I just wrote NON-abs value inequalities, that are equivalent, to represent the fact that {something} is less than 9 units from 0. I always tell students when doing abs value problems like this, "PUT BLINDERS ON!! don't even LOOK at what the STUFF inside the | |'s is!"

Answer 4

@Puffette please respond if you understand what is going on

Answer 5

I like to actually do it one step shorter, so that I don't need to remember which is which.

Answer 6

for me |f(x)| <c becomes -f(x)<c and f(x)<c

Answer 7

let us make it even shorter which is essentially the same, just a notation difference
\[|x| = \sqrt{x^2}\]

Answer 8

You can do that, but it makes it harder.

Answer 9

nope, easier to look at

Answer 10

It increases the power of the inequality, changing a simple line into a quadratic formula and such.

Answer 11

I always change \(\sqrt{x^2}\to |x|\) when simplifying or solving for variables. Undoing that is going backwards. It's really only helpful if you're trying to prove something about the properties of inequalities.

Answer 12

btw, I didn't come up with that idea. I read it in Spivak Calculus 4th ed. like a month ago, and I've been using it since.
My computation is still strictly
|x| = -x ∩ x

Answer 13

yeah for proof only, of course

Answer 14

but that's what we essentially do whenever we introduce a concept, show the proof of why it is so.

Answer 15

By the way, technically it is \(|x|=\sqrt{x\overline{x}}\)

Answer 16

@wio , I didn't follow what you said above, "I like to actually do it one step shorter, so that I don't need to remember which is which. for me |f(x)| <c becomes -f(x)<c and f(x)<c" ??

Answer 17

I mean, I don't understand what is meant by "one step shorter, so that I don't need to remember which is which" lol :)