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Thread: component of composite function: given fcns (f o g)(x) and g(x), how to find f(x) ?

  1. #1

    component of composite function: given fcns (f o g)(x) and g(x), how to find f(x) ?

    so lets suppose i know the composite function : (f o g)(x) and one function g(x).
    How do i calculate the other one? i mean i know that a composite function have several possible pairs of functions,but when i already have one of them stickied it is still many possibilities?

    (f o g)(x) = f(g(x)) I've tried to fill in the functions on the equation but i just get stuck in the f cause this is not a variable like x or y i cant operate with it . I just wanna know if there is an algebric method to find the function or i just need to keep testing differents function that works?
    Last edited by hellawowser; 10-27-2017 at 08:19 AM.

  2. #2
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by hellawowser View Post
    so lets suppose i know the composite function : (f o g)(x) and one function g(x).
    How do i calculate the other one? i mean i know that a composite function have several possible pairs of functions,but when i already have one of them stickied it is still many possibilities?
    I'm sorry, but I don't know what "having (something or other) stickied" means, nor what "it" is that has (or "is"?) "many possibilities"...?

    Quote Originally Posted by hellawowser View Post
    (f o g)(x) = f(g(x)) I've tried to fill in the functions on the equation but i just get stuck in the f cause this is not a variable like x or y i cant operate with it . I just wanna know if there is an algebric method to find the function or i just need to keep testing differents function that works?
    It might help if you provided an example of what you mean, showing what you've tried so far. Thank you!

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    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by hellawowser View Post
    (f o g)(x) = f(g(x))

    I've tried to fill in the functions on the equation but i just get stuck in the f
    Please show us what you were given for fg(x) and for g(x).
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  4. #4
    (f o g)(x) is 2x^2 -4x + 1 and g(x) = 2x - 3. That means f(2x - 3) = 2x^2 -4x + 1 right? How do i get to know the f equation?

  5. #5
    Quote Originally Posted by stapel View Post
    I'm sorry, but I don't know what "having (something or other) stickied" means, nor what "it" is that has (or "is"?) "many possibilities"...?


    It might help if you provided an example of what you mean, showing what you've tried so far. Thank you!
    I've meant that when you only know (f o g)(x) you have several pairs of equation for f(x) and g(x) , but when you know one of them, you still can have a lot of possibilities for the other one?

    Like i said i've only tried to fill in the equation : f(2x - 3) = 2x^2 -4x + 1 but i couldnt operate with it.
    Recently i've tried to analysis the domain and range for each but also couldnt get anything useful from it.
    Last edited by hellawowser; 10-28-2017 at 10:59 PM.

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    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by hellawowser View Post
    (f o g)(x) is 2x^2 -4x + 1 and g(x) = 2x - 3.

    That means f(2x - 3) = 2x^2 -4x + 1 right?
    That's correct.


    How do i get to know the f equation?
    Ask it out for a date?

    One way is to experiment. Start by trying to obtain the second-degree term (2x^2).

    (2x - 3) is the input to function f. What would function f need to do with the 2x part of the input, in order to output a 2x^2 term?
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by hellawowser View Post
    can have a lot of possibilities
    There may be multiple possibilities, but you only need to find one.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  8. #8
    Quote Originally Posted by mmm4444bot View Post
    That's correct.


    Ask it out for a date?

    One way is to experiment. Start by trying to obtain the second-degree term (2x^2).

    (2x - 3) is the input to function f. What would function f need to do with the 2x part of the input, in order to output a 2x^2 term?
    Multiply by x? But in this case wouldnt the x be (2x -3)?

  9. #9
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by hellawowser View Post
    Multiply by x But in this case wouldnt the x be (2x -3)?
    Correct, so let's see what happens, if function f multiplies the input by itself.

    f(x) = x^2

    f(g(x)) = (2x - 3)^2 = 4x^2 plus other stuff

    So, multiplying by the input is not enough. The square of 2x is 4x^2, but we want 2x^2. What to do?

    Answer: Take half the square.

    1/2*(2x)^2 = 2x^2

    So, let's start with that (i.e., make f the function that outputs half of the input's square):

    f(x) = 1/2*x^2

    Now we have:

    f(g(x)) = f(2x - 3) = 2x^2 - 6x + 9/2

    Next step: what can be done about that -6x term, knowing that function f has 2x-3, to work with?
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  10. #10
    Quote Originally Posted by mmm4444bot View Post
    Correct, so let's see what happens, if function f multiplies the input by itself.

    f(x) = x^2

    f(g(x)) = (2x - 3)^2 = 4x^2 plus other stuff

    So, multiplying by the input is not enough. The square of 2x is 4x^2, but we want 2x^2. What to do?

    Answer: Take half the square.

    1/2*(2x)^2 = 2x^2

    So, let's start with that (i.e., make f the function that outputs half of the input's square):

    f(x) = 1/2*x^2

    Now we have:

    f(g(x)) = f(2x - 3) = 2x^2 - 6x + 9/2

    Next step: what can be done about that -6x term, knowing that function f has 2x-3, to work with?

    f(x) = 1/2x^2 + x - 1/2 ??

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