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Compound Functions (Algebra/Fractions)

  1. Mar 15, 2008 #1
    Hello people. This is a problem from an A-Level Math book (though i am independantly learning, have given myself 2 months to cram the whole book down my throat :-S ), so, apparently, it is simple. I understand the principles behind functions, this is more of a problem with understanding complex fractions, i think. Well, here is the question as it is written:

    A function f is defined on the set of real numbers by

    f(x)= [tex]\frac{1-x}{x}[/tex] , (x [tex]\neq[/tex] 0)

    Find, in its simplest form, an expression for f(f(x))

    My Attempt:

    f(f(x)) = f ( [tex]\frac{1-x}{x}[/tex] ) = ( 1-( [tex]\frac{1-x}{x}[/tex] )) / ( [tex]\frac{1-x}{x}[/tex] )

    . . .as you can see, i am rubbish. I know that i am failing to see something. Maybe the following, which will give a further insight to my understanding (or mis), will help someone cure me.

    [tex]\frac{1-x}{1}[/tex] = [tex]\frac{1}{x}[/tex](1-x) = [tex]\frac{1}{x}[/tex] -1

    That is not correct, is it? Where am i going off the track in my understanding?

    One last thing, the text book i am studying has, allegedly, various errors in it, the answer it gives is [tex]\frac{2x-1}{1-x}[/tex] . I can only assume it is right, for now. I just do not see how that answer is arrived at, given the problem.
  2. jcsd
  3. Mar 15, 2008 #2
    You are not going off track anywhere. Your work is correct, it just isn't complete. From what I see, you currently have:

    [tex]f(f(x)) = f\left(\frac{1-x}{x}\right) = \frac{1-\frac{1-x}{x}}{\frac{1-x}{x}}[/tex]

    That's correct, but you do need to simplify the complex fraction, which, as you pointed out, may be where you have a problem. How would you simplify this?

    You noted that (after fixing your typo), that

    [tex]\frac{1-x}{x} = \frac{1}{x}(1-x) = \frac{1}{x} - 1[/tex]

    You can substitute what you found into your complex fraction, replacing all instances of (1-x)/x with 1/x - 1, and after more simplification, you will come up with the right answer.

    The other way to look at the problem, which really is equivalent to what you came up with when you become more familiar with complex fractions, is to combine the numerator and denominator into single fractions. Then, you just need to divide two fractions, which you know how to transform into multiplication.

    The denominator is already a fraction. How would you change the numerator into a single fraction?
  4. Mar 16, 2008 #3
    Thanks, Tedjin. So, i see that

    [tex]f(f(x)) = f\left(\frac{1-x}{x}\right) = \frac{1-\frac{1-x}{x}}{\frac{1-x}{x}}[/tex]

    Multiplying Numerator & Denominator by x would leave [tex]\frac{x-(1-x)}{1-x}[/tex] = [tex]\frac{x+x-1}{1-x}[/tex] = [tex]\frac{2x-1}{1-x}[/tex]

    Which may not seem to have been the method i was heading towards. Last night, (1-x)/x = (1/x)(1-x) = (1/x)-1 , seemed logical to me, but now, whatever logic i used to go from (1-x)/x to (1/x)(1-x) has escaped me. . i see that it is correct, in as far as that given a value for x, it results in the same answer. .but it seems more logical now to go from (1-x)/x and multiply by x, leaving x(1-x) = x-x[tex]^{2}[/tex] ..

    I will be back later, but have some work to do for now. Thanks again. Anything else from you or anyone which could be helpful would be lovely.
  5. Mar 16, 2008 #4
    Right, you've got it. Multiplying by 1 to clear out the fractions in the numerator and denominator is a very useful way of simplifying complex fractions.

    Now, you are right that (1-x)/x = (1/x)(1-x), because dividing by x is the same thing as multiplying by 1/x. Then, of course, you have the distributive property. One thing you should watch out for in your post is that (1-x)/x * x = 1-x. It is not equal to x(1-x). It seems like you know your algebra; you just need to be more careful.
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