Well still I am having problem to understand it...

As function tends to +/- infinity at x  = +/- 3 we should look at 5 cases

(-∞,-3)     (-3,0)   0    (0,3)    (3, ∞)

How you can take the case (-∞,-3) as x can't be -4 or 4 ?

I did it a different way I just assumed two cases when x>= 0 and x<0
Here is what I am doing

So we can write

This far I did it... and I can also see that x can't be -3 or 3 as it will make the denominator 0.
But I can't find a proper steps to come to a proper solution set...I am still trying....

Last edited by debjit625 (2012-11-23 19:35:02)

Q1.

Look at the graph I posted.  You can see that + / - 4 is calculable and that values between -3 and + 3 satisfy the inequality.

So for example

-3 <  x < 0

Now in that range for x,  (-x-3) is negative

eg. (--2.9 - 3) = -0.1

So when you multiply by that denominator you must reverse the inequality:

This final statement is true in that range so the range is part of the solution set.

I'll look at the other questions once you are happy with this one.

Bob

Bob

HI debjit625

Sorry, I just misunderstood what you were saying about + / - 4

What you have just said is correct, no problem.

In post 1 you had two ranges, in words below -5 and above + 5

Those are both correct and my graph shows these.

What was missing from your answer was another range of possible values for x, between -3 and + 3

Look at my graph and you can see that, in this range, 'y' is always negative, so certainly < 1/2

So why did these values 'go missing' ?

The graph has a discontinuity at -3 and again at +3 because of the division by zero.

So you need to check in this range too.

I tried to demonstrate why (-3,0) is another set of values that fit the inequality.  Twice I had to use the rule that 'if you multiply or divide by a negative, you must reverse the inequality'.  Do you know and understand this rule?

I have recently posted an explanation at

[url]http://www.mathisfunforum.com/viewtopic.php?id=18411[/math]

You might like to have a look at that, especially post 7 in that thread.

I've got to log off for a short while.  I'll be back on in about 15 mins.

ADDITIONAL EDIT:

case 3 which you actually explained in post #5 gives

but as already given

If I call

... statement one and   

... statement two, then we can work the logic like this.

I got statement one by algebra assuming statement two.  So statement one is only true to the extent that it obeys statement two.  As statement two is a subset of statement one that means the inequality is satisfied just for the subset ie

So that provides part of the missing answer. 

Testing x = 0 by substitution shows it may be added to the set giving

You can finish by considering (0,+3) and showing it is legitimate to add this to the range
Bob

hi debjit625

Let's move on to Q2 and maybe it will become clearer.

x + 2 changes sign  at x = -2 so that will divide the cases we need to consider.

There's also something to consider at x = 0, as the expression goes to infinity there.  It looks as though this will go to - infinity on one side of zero and + infinity on the other.

To simplify the expression you will have to multiply by x, and that change of sign will influence the inequality sign.

So you need to consider 4 cases:

(i) x < -2

(ii) x = -2

(iii) -2 < x < 0

(iv) 0 < x

Do you want to try this and post what you get or would you like more help with this ?

Bob

@ bob bundy
Ok I did all the problems,but I have some question regrding post #9 where you posted about question 2.

There's also something to consider at x = 0, as the expression goes to infinity there.  It looks as though this will go to - infinity on one side of zero and + infinity on the other.

Well how can  x = 0 makes the expression goes to +/- infinity as dividing anything by 0 is undefined,and x is the denominator.

What solution set did you got for question 2 , 3 and 4?

Thanks

Well I was a bit busy so I couldn't post...

@bob bundy ,thanks I have also got the same answer.