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Express as a Single Simplified Fraction

  1. Sep 14, 2008 #1

    thrill3rnit3

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    1. The problem statement, all variables and given/known data

    sorry I don't know how to type algebraic expressions, but I'll do my best.

    express quantity [ f(x+h) - f(x) ] over h as a single simplified fraction

    f is not a variable, it's a function

    2. Relevant equations

    Question # 1. function of x = quantity 1-x over x

    3. The attempt at a solution

    umm I don't need answers. I just need to know what to do first. Do I substitute the given function of x in the given equation? Do I add the h to the given function? Like say for number 1, do I do (quantity of 1-x over x) + h??

    Sorry I'm asking a noob-ish question. I'm not on my best thinking mode right now, especially that things aren't going my way lately...

    Thanks :D
     
    Last edited: Sep 14, 2008
  2. jcsd
  3. Sep 14, 2008 #2

    danago

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    So you are given that [tex]f(x)=\frac{1-x}{x}[/tex]?

    To find f(x+h), all you do is replace any 'x' you see with an 'x+h', so:

    [tex]f(x+h)=\frac{1-(x+h)}{x+h}[/tex]
     
  4. Sep 14, 2008 #3
    do you mean [tex]f(x) = \frac{1-x}{x}[/tex] and you need to calculate f(x + h) - f(x).

    f(x + h) means that in the given function you must replace 'x' by 'x + h' and NOT adding h to the ENTIRE function
     
  5. Sep 14, 2008 #4

    thrill3rnit3

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    oh I see, I see. That answers my question. Thanks a lot guys :cool:
     
  6. Sep 14, 2008 #5

    thrill3rnit3

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    so for the first problem, is this correct:

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

    plus further simplification, of course.
     
  7. Sep 14, 2008 #6

    Dick

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    That's exactly it.
     
  8. Sep 14, 2008 #7

    thrill3rnit3

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    How about [tex]\frac{1}{x^{2}}[/tex] ? Should I replace [tex]f(x+h)[/tex] with [tex]\frac{1}{x^{2}+h}[/tex] ??
     
  9. Sep 15, 2008 #8

    Dick

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    No. If f(x)=1/x^2 then f(x+h)=1/(x+h)^2. I thought you had this.
     
  10. Sep 15, 2008 #9

    thrill3rnit3

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    Yep. Thanks.
     
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