Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Math Help:Reciprocal & Graphing

  1. Dec 10, 2004 #1
    What would be the reciprocal of the following:

    [tex]A)y=x^2[/tex]

    [tex]B)y=\sqrt{x}[/tex]

    I think the reciprocal for A, it would be:

    [tex]y=\frac{1}{x^2}[/tex]

    And B would be:

    [tex]y=\sqrt{\frac{1}{x}}[/tex]

    And also, How would I graph this?
    I just need the table of values. Well, I need to do domain,range,graphing and a summary but I think I can do that myself if I get how to do the table of values.


    Thanks :smile:
     
  2. jcsd
  3. Dec 11, 2004 #2

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you are talking about reciprocals of numbers, then yes, the reciprocals of [itex]x^2~and~`\sqrt(x)~are~\frac{1}{x^2}~and~\frac{1}{\sqrt(x)}[/itex]

    Since no domain has been provided in the question, you have to come up with what you think will be a relavant domain- ie : one which includes any "important" behavior.

    For [itex]y = x^2 [/itex], I would pick a bunch of numbers say, from the set [-10, 10].
     
  4. Dec 11, 2004 #3
    oops - lol nvm.
     
  5. Dec 11, 2004 #4
    Would this be the graph:
    GraphReciprocal.jpg

    The blue one is [tex]\frac{1}{x^2}[/tex] and the red one is [tex]\sqrt{\frac{1}{x}}[/tex]
     
  6. Dec 11, 2004 #5

    HallsofIvy

    User Avatar
    Science Advisor

    One question: since if f(x)= x2 (for x>= 0), then f-1(x)= &sqrt;(x) are the functions you give, are you sure it is "reciprocal" that is intended rather than the inverse function?
     
  7. Dec 11, 2004 #6
    I don't get your question. Please rephrase it.
     
  8. Dec 11, 2004 #7
    He means are you sure you're meant to be doing 1/y or should you be finding the inverse function (eg. the inverse of multiplication is division, the inverse of squaring is square rooting)?
     
  9. Dec 11, 2004 #8
    I am sure that I don't have to find the inverse function.
     
    Last edited: Dec 11, 2004
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook