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Sum1 here help

  1. Sep 13, 2003 #1
    sum1 here help!!!

    ok....since no1s helpin in the hmwk section thot id take a shot here..

    if f(x)= (2e^x -8)/(10e^x + 9)
    then wat is f^(-1)(x)?

    i first took the ln of both sides....

    getting lny = ln (2e^x - 8)/(10e^x + 9)

    then using one of the properties i get

    lny = ln (2e^x - 8) - ln (10e^x + 9)

    and from here i get stuck...how do i solve for x?? am i doing it a totally wrong way?? plz help!!!!!

    and i did get a sol'n from sum1 but i had a question abt it.......some1 b kind enuf 2 help!!!
     
  2. jcsd
  3. Sep 13, 2003 #2
    f^-1 = 1/f

    1/((2e^x -8)/(10e^x + 9)) = (10e^x + 9)/(2e^x -8)

    then you can work to write this function simpler.
     
  4. Sep 13, 2003 #3
    f-1 is the inverse of f, where f is a function.

    f-1 does not equal 1/f
     
  5. Sep 13, 2003 #4
    if x^-1 = 1/x
    it would mean the same for a function, so

    f^-1 would be the same as 1/f
     
  6. Sep 13, 2003 #5
    Sorry, you're wrong Astrophysics. If f is a function then f^-1 is the inverse of the function so that f^-1(f(x)) = x.
     
    Last edited by a moderator: Sep 13, 2003
  7. Sep 13, 2003 #6
    Re: sum1 here help!!!

    You just need some algebraic manipulation first

    f = (2e^x -8)/(10e^x + 9)
    so, 10e^x * f + 9f = 2e^x -8
    so, 10e^x * f - 2e^x = -8 - 9f
    so, e^x ( 10f - 2 ) = -( 8 + 9f )
    so, e^x = -( 8 + 9f ) / ( 10f - 2 )
    so, x = ln[ ( 8 + 9f ) / ( 2 - 10f ) ]
    so, x = ln( 8 + 9f ) - ln( 2 - 10f )

    therefore f^-1(x) = ln( 8 + 9x ) - ln( 2 - 10x )
     
  8. Sep 13, 2003 #7

    oh thanks mathnerd, but...how come when I put this in my calculator I get f^-1 = 1/f and f^-1(f(x)) is not x??????
     
  9. Sep 13, 2003 #8
    Because your calculator thinks you're talking about the multiplicative inverse. It doesn't realize that you really want the inverse of the function. In mathematical notation f^-1(f(x)) = f(f^-1(x)) = x e.g. cos^-1 x does not equal 1 / cos x, it is instead inverse cos which is the inverse function of the cosine function. In my last post where I solved for f^-1(x), you can easily show that f(f^-1(x)) = f^-1(f(x)) = x
     
  10. Sep 13, 2003 #9
    I'm very confused here, I was actually thinking the same thing as my calculator I guess, but I have no idea what these inverse means, is it f^-1 or is it something like f' ???
     
  11. Sep 13, 2003 #10
    or you could say cos^-1 x=arccos x
     
  12. Sep 13, 2003 #11
    Yes, well ArcCos is nothing but the inverse of the cosine function, whereas 1/cos(x) = sec(x), it is easily shown that sec(x) doesn't equal ArcCos(x). By definition the inverse of an arbitary function f(x) is another function g(x) such that g(f(x)) = f(g(x)) = x. In mathematical notation g(x) = f^-1(x)
     
    Last edited by a moderator: Sep 13, 2003
  13. Sep 13, 2003 #12
    inverses and identities.

    unification.

    multiplication.
    1 is the "multiplicative identity." let x be a nonzero real number.

    x * x^-1 = x^-1 * x = 1. in effect, the thing and its inverse cancel, leaving the identity.

    functions.
    let * stand for the composition of functions. so rather than write
    f(g(x)), write f * g (x).

    f * f^-1 = f^-1 * f = I. by, "I", i mean the "identity function,"
    I(x) = x.

    except for zero, all numbers have a mulitplicative inverse. not so for functions. for example, the function f(x) = x^2 has no inverse defined on the set of real numbers. however, if restriced to nonnegative numbers, f does have an inverse, namely the square root of x. that is precisely what will "cancel" or "oppose" the operation of squaring.

    cheers,
    phoenix
     
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