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Addition and Multiplication of functions with radicals in the denominator

  1. Aug 2, 2011 #1
    1. The problem statement, all variables and given/known data
    f(x)= 1/(5x2 +3)
    g(x)=1/(x-2)1/2

    find (f+g)(x)


    3. The attempt at a solution

    for (f+g)(x) I use the common denominator by multiplying each side by (x-2)1/2/(x-2)1/2 and (5x2 +3/5x2 +3) respectively.

    I end up with a whole mess that I am unable to rearrange into anything useful, the (x-2)1/2 completely throws me off my game.

    Any ideas?
     
  2. jcsd
  3. Aug 2, 2011 #2
    [tex] \frac{\sqrt{x-2}}{(\sqrt{x-2})(5x^2+3)}+\frac{5x^2+3}{(\sqrt{x-2})(5x^2+3)}= [/tex]
    [tex] \frac{(\sqrt{x-2})+(5x^2+3)}{(\sqrt{x-2})(5x^2+3)}
    [/tex]
    Isn't that what it is?
     
  4. Aug 3, 2011 #3
    Same answer as me, I guess we can't be both wrong.

    How about (f x g)(x)?
     
  5. Aug 3, 2011 #4

    eumyang

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    Homework Helper

    Why don't you show us, and then we'll let you know if it is correct.
     
  6. Aug 3, 2011 #5
    Thanks for the help, for some reason I thought I could further simplify that.

    So for (f x g)(x)

    It'd be

    1/(5x2+3) x 1/(x-2)1/2=

    1/(5x2)*(x-2)1/2 + 1/3((x-2)1/2

    That's where my problem lies, I have forgotten how to deal with the root of x-2 when trying to simplify this further. Also, as I'm new to the site, could one of you point me to the equation editor you guys use in your posts? It makes everything look a lot neater and easier to understand.
     
  7. Aug 3, 2011 #6

    eumyang

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    Homework Helper

    You don't need to do that last step. What you had previous to that is sufficient:
    [tex]\frac{1}{\sqrt{x - 2}(5x^2 + 3)}[/tex]
     
  8. Aug 3, 2011 #7
    We type it using Latex. You should be able to see the original text if you hover or right click I think. You start with [ tex ] (without the spaces) and end with [/ tex] use itex if you want the equations to appear in line with regular text.
     
  9. Aug 3, 2011 #8
    You guys have been great help, thank you all very much. I'll try to contribute towards answers as well once I get back in gear!
     
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