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