"Rules" for composite functions

• donkhead333

  9 Dec 09, 11:33

  For the JC topic..

  I have only learnt thus far that for the composite function, gf, to exist, it must comply with at least 1 of these rules:

  1)Rf subset of/equals Dg

  2)Dgf equals Df

  3)Rgf subset of/equals Rg.

  Then I encountered this statement in one of the answer keys:

  Rf^(-1) subset of/equals Dg.

  Could anyone explain this to me? Thanks!

• wee_ws

  9 Dec 09, 16:54

  Hi, 

    Only point 1 is necessary for the existence of gf.

    Point 2 helps us to define the domain of gf. 

    To find the range of gf, it will be helpful to draw the graphs of f and g.

    As for your enquiry, please provide the original question for the sake of clarity.

    Thanks.

  Cheers,
  Wen Shih  

• donkhead333

  9 Dec 09, 18:44

  Opps, posting the full question completely slipped my mind, it is pretty long actually

  f = 3x^2 - 6x - 1 = 3(x - 1)^2 - 4

  Rf = [-4, infin.) ~ (by the way, why is the -4 inclusive? wont it y=k cut at 2 points at y=-4?)

  for f^-1 to exist, x must be equals to or greater than 1.

  f^-1 = sqrt [(x+4)/3]+1  ---x equals to or greater than -4 (same question as abovementioned point)

  function g = sqrt(x+4), x equals to or greater than -4

  Determine whether the composite function gf^-1 exists. Find the range of gf^-1, if it exists.

• wee_ws

  10 Dec 09, 08:17

  Originally posted by donkhead333:

  Opps, posting the full question completely slipped my mind, it is pretty long actually

  f = 3x^2 - 6x - 1 = 3(x - 1)^2 - 4

  Rf = [-4, infin.) ~ (by the way, why is the -4 inclusive? wont it y=k cut at 2 points at y=-4?)

  for f^-1 to exist, x must be equals to or greater than 1.

  f^-1 = sqrt [(x+4)/3]+1  ---x equals to or greater than -4 (same question as abovementioned point)

  function g = sqrt(x+4), x equals to or greater than -4

  Determine whether the composite function gf^-1 exists. Find the range of gf^-1, if it exists.

  Hi,

    Is the domain of f given in the question?

    Assuming that the domain of f is the entire real set of numbers, the range you stated makes sense. At y = -4, the horizontal line is tangent to the curve of f.

    Yes, you are correct that x >= 1 for the inverse of f to exist. Another possible maximal domain is given by the set of x such that x <= 1.

    To check for the existence of the composite function g f^(-1), we consider range of f^(-1) and domain of g:

    Range of f^(-1) = domain of f = (1, inf);

    Domain of g = (-4, inf), which has been given.

    Now (1, inf) is indeed a subset of (-4, inf), so the composite function g f^(-1) does exist.

    To find the range of g f^(-1), we take these systematic steps:

    1. Sketch the graph of y = g(x), a C-shaped half-parabola.

    2. Map the range of f^(-1) onto the x-values in the graph.

    3. Find the set of y-values corresponding to point 2.

    So the range of g f^(-1) = ( sqrt(5), inf ).

    Thanks.

  P.S. Do give the question "as it is" in future. Thanks.

  Cheers,
  Wen Shih