absolute value of (2 + 1/x) > 1

[roadrunner]

Here is the problem:

absolute value of (2 + 1/x) > 1

I can solve this by finding the critical points 0, (x cannot = 0), -1 and - 1/3 (these by solving the problem as an equality). After finding the critical points I tested a point in each interval to determine the correct intervals. I understand this.

I am having trouble solving this problem using cases.

Any ideas? I can't seem to get the same answers when I try to use cases. I tried x > 0 and x < 0 etc.

Thanks for any help you can give.[/list]

 

[tkhunny]

Re: absolute value

|2 + 1/x| > 1

Have you tried solving this

(2 + 1/x) > 1

OR this

(2 + 1/x) < -1

??

Where did I get those?

 

[roadrunner]

absolute value of (2+ 1/x ) > 1

Hi and thanks for your replies.

Yes, thkunny, I know how to solve an absolute value problem using cases. I did originally set up my work as you suggest. But I was unable to get the correct intervals. I have checked for sign errors, etc. I must be missing something very simple.

Once I get to 1/x>-1 or 1/x < -3 do I then have to do cases for when x > 0 and when x < 0? Because multiplying by a an x with a negative value changes the sign, etc.

For 2 + 1/x > 1 letting x > 0 I get x > -1 , letting x < 0 I get x < 1.

For 2 + 1/x < -1 letting x > 0 I get x < -1/3, letting x < 0 I get x > 1/3.

Are these right and I just don't know how to use them? If not right can you point out the error(s).

The second suggestion by galactus includes 0 in the interval (-1/3,infinity) which is not correct. But how did you get these intervals, the same way I got them using critical points?

 

[galactus]

May I ask how 0 is not correct?. It is > 1 at 0.

Look at the graph. That should help you see.

 

[tkhunny]

Are you sure you're not on illegal substances, galactus? x = 0 isn't in the Domain. However, it is a nice, convincing picture.

|2 + 1/x| > 1

Okay, what I REALLY would do is forget anything fancy or convenient and just beat it up as thoroughly as possible.

First, you have to buy into this concept:

For x >= 0, |x| = x Example: |3| = 3
For x < 0, |x| = -x Example: |-2| = -(-2) = 2

Solve 2 + 1/x = 0.

2x + 1 = 0
2x = -1
x = -1/2

We already have the problem at x = 0, so this definesd THREE regions about which we must be concerned.

x <= -1/2 -- 2 + 1/x >= 0 here.

-1/2 < x < 0 -- 2 + 1/x < 0 here.

0 < x -- 2 + 1/x > 0 here.

So,

For x <= -1/2 or x > 0, we have 2 + 1/x > 1 to solve.

For -1/2 < x < 0, we have -(2 + 1/x) > 1 to solve.

Solving the first:

For x <= -1/2
2 + 1/x > 1

Multiply by x that we know is negative
2x + 1 < x
finish up.
x < -1 <== There is a chunk of the solution, since all of it is less than -1/2.

For x > 0
2 + 1/x > 1

Multiply by x that we know is positive
2x + 1 > x
finish up.
x > -1 <== This is NOT a chunk of the solution, since it applies ONLY for:

For x > 0 <== There is a chunk of the solution.

Folving the second piece:

For -1/2 < x < 0
-(2 + 1/x) > 1
-2 - 1/x > 1

Multiply by x that we know is negative
-2x -1 < x
-1 < 3x
-1/3 < x <== There is a chunk of the solution, remembering that is apples ONLY for -1/2 < x < 0.

Finally, the solution:

|2 + 1/x| > 1 for
x < -1 or x > 0 or -1/3 < x < 0

It wasn't pretty, but with exceptional care, one can wade through it. galactus's picture is helpful, but I'd have to include the line y = 1, and maybe zoom in to -2 < x < 1 and -2 < y < 2 really to see the circumstances.

 

[galactus]

As a matter of fact, I am on legal and illegal substances.

That's why one should never drink and derive.

A picture is worth a 1000 words. I say that's the way it is and I don't care what anyone says(puts fingers in ears and hums real loud) :lol:

Just kidding. :wink: :lol:

I had originally posted it as \(\displaystyle \L\\(-\infty,-1)\cup(\frac{-1}{3},0)\cap(0,\infty)\)

But thought I made a discombobulation.

I see the interval is in the \(\displaystyle open(-\infty,-1),open(\frac{-1}{3},0),open(0,\infty)\)

If I am still misconstrued, the h*ll with it. :roll: [/quote]

 

[roadrunner]

Thanks for being able to laugh at yourself, galactus.

Tkhunny, I can follow your explanation. However, I think my concept of using the three critical points and then testing the intervals is quickest and easier and is a concept that I was taught in calculus.

I don't think we get enough problems like these to help us improve our skill.

I'm sure I'll be back for something else.

I really appreciate your 'service'. I think that because you won't just do a problem is great. Submitters should demonstrate what they have accomplished to you before you spend time helping them.

:!: THANK YOU

 

[jonboy]

If I am still misconstrued, the h*ll with it. :roll:

LOL I feel this way sometimes.