Expert: Scotto - 7/16/2008

Question
QUESTION: 1) f(x)= (2x-3)/(x-5)

What is modulusf(x)modules > (or equal to) 3/2.

I have found the answer of f(x) > (or equal to) 3/2 to be x< (or equal to) -9 or x> 5 and am sure this is correct. However, I am not sure how to solve the qn above.

2) Given a,b are positive constants and a<b, solve for x:

(x-b)/(x-a) > (x-b)/b

Thanks for the help sir :)



ANSWER: For future references, ≥ is alt-242 and ≤ is alt-243.

1) I'm not sure what modulusf(x)modules are.

2) Note that the answer needs x=a generates an error.  IN the langues C, this is referred to as x!=a (x is not equal to a).
The (x-b) can be cancelled of each side.
Inverting, this leaves x-a≥b, or x≥a+b.


---------- FOLLOW-UP ----------

QUESTION: 1) I meant modulus(fx)...

2) the correct answer shld be x<a or b<x<a+b, which I cannot seem to arrive at..

would you be able to help out with these 2 qns again pls..

thanks :)

ANSWER: Sorry for the error in my answer to 2.  It should have read:
2) Note that the answer needs x=a generates an error.  In the computer language C, this is referred to as x!=a (x is not equal to a).

Now to your current question:
1) When the modulus of a function is taken, it is generally shortened as 'mod'.  An example would be like
7 mod 3 = 1 or
12 mod 5 = 2.

When you are given a mod b, the result is the remainder when dividing a by b.

I still don't know what modulus[f(x)] refers to since there is no term b in the expression.  If it was modulus [f(x),2], that would be the remainder on f(x) when dividing by 2 for each x.

An example would be f(x)=2x and we look at mod[f(x),3].  This would generate a series of lines starting at 0 and going up to 3 on an infinite number of intervals between -∞ and ∞.  They would be ... (-1.5, 0), (0, 1.5), (1.5, 3), (3, 4.5) ... .

2) We are looking at (x-b)/(x-a) > (x-b)/b, a<b.
Depending on the values of a and b, I get different results and can’t arrive at that answer either.

When x>a and b<0, the equation is true.
When x<a and b>0, the equation is true.
When x<a+b with (x<a, b<0), it is true.
When x<a+b with (x>a, b>0), it is true.

I designed the entire problem in a spreadsheet and found it to be true in a variety of places.  The critical lines were b=0 and a=x.  In each of the regions, the function turned out to be all true, all false, or a mixture with cutting lines on diagonals.  I did get it to always be true if a>x.

The rest, though, is more than I can get an answer for right now.


---------- FOLLOW-UP ----------

QUESTION: thanks scotto...i have a clarification for the modulus..

1) it is interesting to remember that 12 mod 5 is 2. The way I was looking at the original problem I gave you, my interpretation of mod was that of a sketch, in which the negative region is reflected about an axis into the positive region..

for example...f(x)=x, thus mod f(x)= -x (for x<0) and x (for x>0)...in a graph, we would draw a V where the base of the V touches the origin..

Hence, based on that...I was trying to solve the original qn.. Would there be a way to do this?

2)I am truly thankful you took the effort to do up this in a spreadsheet.. cld I have a look at it? If you used google docs, that'd be cool, tho i can MS office too, so either way shld be good..

Thanks!

Cheers,
baroque_87

Answer
1) If f(x)=(2x-3)/(x-5), then by what you have said the mod looks like it takes the absolute value of f(x).

What needs to be done is to find where |(f(x)|>3/2.

2) I have the spreadsheet all set up for you.  It has four different tables in it.  

They are Main, a<b, (x-b) over (x-a), and (x-b) over b.

The tables a<b, (x-b) over (x-a), and (x-b) over b have in them exactly what they are called.

In the first one are the final results with the variables in A1:A3.
A description of the values are in B1:B3.

I need someway to send this file to you, since trying to attatch it as an image file won't work.  It's a great looking file.  Note that when you do get it, various values can be tried in all three of the cells A1:A3.

The file is on my computer as Inequality.xls in \AllExperts_files, but I'm not sure how to sent it yet.