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Reciprocal functions ?

  1. Sep 28, 2007 #1

    JPC

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    reciprocal functions ??

    hey

    is this true with finding reciprocal functions :

    f(x) initial function , g(x) reciprocal function of f(x)

    if f(x) = a x^n
    then
    g(x) = ( xroot(n, x) ) / ( xroot(n, a)

    but now
    if f(x) = ax²+ bx + c

    how do i find its reciprocal function ?
     
  2. jcsd
  3. Sep 28, 2007 #2

    EnumaElish

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    By reciprocal do you mean inverse, defined as "g is the inverse function of f if f(g(x)) = x"?
     
  4. Sep 28, 2007 #3

    JPC

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    yes i think, i mean like :
    if f(x) = Y

    then g(Y) = x

    (sorry if its the wrong vocab, but im in a french school)
     
  5. Sep 28, 2007 #4

    EnumaElish

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    Okay, in English it's "inverse function," written f-1.

    How do you solve for f-1? If you have y = (an expression with x and no y) then your goal is to obtain x = (an expression with y and no x). For this to work, it is necessary to isolate x. For example, if you have y = ax^2 + bx + c then you need to turn this into an equation with 0 = ax^2 + bx + c - y. Now, what is a method to solve this equation, so that x = (an expression with a, b, c, y)?
     
  6. Sep 29, 2007 #5

    Gib Z

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    EnumaElish was on the right track, but all he is doing was writing the same function, but with the subject being x instead of y. For the inverse or reciprocal function, replace all the x's with y's and vice versa, then make the subject y, which is done with EnumaElish's hint.
     
  7. Sep 29, 2007 #6

    HallsofIvy

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    Some people, my self and apparently Gib Z, prefer to first swap x and y, and solve for y. Others, like EunumaElish solve for x first, then swap x and y. It's six of one, half dozen of the other.
     
  8. Sep 29, 2007 #7

    Gib Z

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    O Sorry it's just that from EnumaElish's post i thought that he wasn't going to end up swapping the x and y's. If he was, then it would be correct.
     
    Last edited: Sep 29, 2007
  9. Sep 29, 2007 #8

    EnumaElish

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    The way I was taught was to first turn the equation around, then re-label.
     
  10. Sep 29, 2007 #9

    JPC

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    yes buyt how do i do it after :

    ax²+ bx + c = y

    ax² + bx = y - c

    i need to get it into the form of x = ...... (with y on this side)
     
  11. Sep 29, 2007 #10

    EnumaElish

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    Correct. And you need an individual x on the left. So, how do you express individual x when your equation is a quadratic polynomial?
     
  12. Sep 29, 2007 #11

    HallsofIvy

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    Remember that only one-to-one functions have inverses!

    You can solve ax²+ bx + c = y using the quadratic formula- that typically gives two values because quadratics are typically not one-to-one functions.

    To take an easy example, if y= x^2 then [itex]x= \pm \sqrt{y}[/itex]- which is not a function. When we "really, really" want to have an inverse, what we normally do is create two new functions f(x)= x2 for [itex]x\ge 0[/itex] and g(x)= x2 for [itex]x\le 0[/itex]. The inverse of f is [itex]f^{-1}(x)= \sqrt{x}[/itex] and the inverse of g is [itex]g^{-1}(x)= -\sqrt{x}[/itex].
     
  13. Sep 29, 2007 #12

    Gib Z

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    (Continuing from Hall's Post)

    Or you can limit the domain of the quadratic function so that it is one to one in the new domain, and define its inverse function only from that domain.
     
  14. Sep 29, 2007 #13
    try to complete the square
    if [tex]x^2 + 2 a x = y[/tex]
    then we can do [tex] x ^2 + 2 a x + a^2 = y + a^2; (x+a)^2 = y + a^2[/tex]
    you should be able to go on after this. Just one little caution then you are done
     
  15. Oct 4, 2007 #14
    Leon1127,
    nice job there, but [tex](x+a)^2 = y + a^2[/tex] doesn't solve for x. How do you derive x from that equation? Are you still considering the coef b in your equation?
     
    Last edited: Oct 4, 2007
  16. Oct 4, 2007 #15

    Gib Z

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    Square root and subtract a from both sides.
     
  17. Oct 5, 2007 #16

    JPC

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    but where have u put the rest ; b and c :

    i mean reciprocal functions for functions of the type : y= ax² + bx + c

    so if we replace :

    x= ay² + by + c
    ay²+ by = c - x

    but now i dont know how i can get y on the form of y = ...
     
  18. Oct 5, 2007 #17

    EnumaElish

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    Can you solve for y in ay²+ by + d = 0 where d = -(c - x) ?
     
  19. Oct 5, 2007 #18
    Yeah. I solved it last nite. Thanks guys. This was the only forum on the entire net I found a hint to solve this problem!! keep up the good work.
     
  20. Oct 5, 2007 #19

    JPC

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    oh yes
    thanks
     
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