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Topic review (newest first)
- noelevans
- 2012-11-19 16:53:53
Hi! 
Those of you who have access to a good graphing program might try graphing the function
f(x) = 3|x-2|/x and then draw the horizontal line y=4 and see where the graph lies below
or on that line. I think you may find that any x in (-inf, 0) union [6/7, inf) will have its output
less than or equal to 4. (I graphed it by hand so that could be a "bit shaky". I'd like to see
the graph of this function using a good graphing program. Hint! Hint!)
I'm still a bit confused about what the original question is. The domain of the function above
is any real number except zero. The range seems to be the set R-[-3, 0), R being the set of reals.
If we require f(x) <= 4 then this is true for a subset of the domain, namely that mentioned above:
(-inf,0) union [6/7, inf). Any x in the subset [1, inf) of the domain will certainly satisfy f(x) <= 4.
If we require that f(x) be in [-3, 0) then the domain is restricted to the empty set.
Any restriction on the outputs of f(x) corresponds to a subset of the domain R-{0}.
Solving the original inequality is essentially finding out what subset of the domain of f(x) will make
the inequality true. To do so I usually consider all possible cases that make a difference in the
expressions replacing the absolute value expressions. It's a bit tedious, but I tend to get lost if I
don't consider each case. The union of the sets obtained in each case is typically the final answer.
Also in each case it is the intersection of the assumed condition and its result that gives the set for
that case.
Determining the range of the function is a different matter altogether.
Note: We could also make the function f(x) = (3|x-2|/x) - 4 and see where it is <= zero.
Absolute value inequalities are tricky. It is probably best to write an appropriate function f(x)
and graph it to see what the solutions are keeping in mind that the solution set is a subset of
the domain of the function.
GOTTA GET SOME 
- anonimnystefy
- 2012-11-19 13:56:55
First we need to split the problem into two cases.
The first case is when x is in [1,2). In that case the equation becomes 3*(x-2)/x<=4 which then comes down to x>=-6. Since this is true for x in the said interval, we can move on to the second case, which is x>=2. In this case the equation becomes 3*(2-x)/x<=4 and reduces to x>=6/7, which is true for x>=2. Since both cases are satisfied the proposition holds.
- bobbym
- 2012-11-19 13:47:48
Hi;
That I am not getting at all. Care to explain it?
- anonimnystefy
- 2012-11-19 13:39:33
No, no. The inequality comes down to x>=6/7 only for x>=2. To prove it for 1<=x<2 you have to use the case where you get x>=-6.
- bobbym
- 2012-11-19 13:37:10
Hi;
I am not following you. Is it not true that it is true for x >= 6 / 7? Does that not imply greater than 1 also?
- anonimnystefy
- 2012-11-19 13:19:12
The first case is for reals in [1,2) because that is where x-2 is positive. Only for larger numbers should the second case be used.
- bobbym
- 2012-11-19 13:14:42
Hi;
Why two? We are not talking about integers.
- anonimnystefy
- 2012-11-19 13:12:20
bobbym wrote:
Hi;
There is a minor flaw in this proof. You are using the second case for all x>=1, when you should be using it for x in [2,infty).
- bobbym
- 2012-11-19 13:07:23
Hi;
I hope it is what the OP wanted, but he never came back. I do not know if that is good or bad.
- daydoe12
- 2012-11-19 12:59:54
ok that makes sense
- bobbym
- 2012-11-19 12:58:28
Hi;
It proves it for all real numbers greater then 6 / 7.
- daydoe12
- 2012-11-19 12:56:32
how is that proving for all real numbers
- bobbym
- 2012-11-19 12:48:56
Hi;
See if you can use it.
- daydoe12
- 2012-11-19 12:44:42
no but i will thanks
- bobbym
- 2012-11-19 12:43:01
Hi;
Did you click the hidden text in post #6?