AboutScotto Expertise Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.).
I also have answered some questions in
Physics (mass, momentum, falling bodies),
Chemistry (charge, reactions, symbols, molecules), and
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Experience Experience in the area: I have tutored students in all areas of mathematics for over 20 years.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
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That's around 2,000 in basic math and 1,000 in advanced math.
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I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few junior level courses my sophomore year.
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All of my scores right now are at least a 9.8 average (out of 10).
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referals from a company, friends and aquantenances, people from my church, and people like you.
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.
