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Math Help - Constructions of functions

  1. #1
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    Constructions of functions

    So here's the question:
    Find fg (relation) for each pair of functions f and g. Use understood domains for f and g.

    a.) f(x)=2x+5, g(x)=6-7x

    (fg)(x)= 2(6-7x)+5 = 17-14x

    (gf)(x)= 6+7(2x+5) = -14x-29

    Okay i understand this. But

    b.) f(x)={(t,r),(s,r),(k,l)} , g(x)={(k,s),(t,s),(s,k)}

    fg = {(k,r),(t,r),(s,l)} but gf= (null set)

    i see how fg works, replace the x value of f with the x value of g. But why is the relation of gf the null set?
    wouldnt it be {(t,s),(s,s),(k,k)}?
    Now i see that t and s have the same y variable 's' which doesent make this a function.
    BUT f(x) has t and s with the same value 'r' so its not a function in the first place right??

    Please help a fellow understand. Pictures would be great.
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  2. #2
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    Re: Constructions of functions

    Quote Originally Posted by Aquameatwad View Post
    So here's the question:
    Find fg (relation) for each pair of functions f and g. Use understood domains for f and g.

    a.) f(x)=2x+5, g(x)=6-7x

    (fg)(x)= 2(6-7x)+5 = 17-14x

    (gf)(x)= 6+7(2x+5) = -14x-29

    Okay i understand this. But

    b.) f(x)={(t,r),(s,r),(k,l)} , g(x)={(k,s),(t,s),(s,k)}

    fg = {(k,r),(t,r),(s,l)} but gf= (null set)

    i see how fg works, replace the x value of f with the x value of g. But why is the relation of gf the null set?
    wouldnt it be {(t,s),(s,s),(k,k)}?
    Now i see that t and s have the same y variable 's' which doesent make this a function.
    BUT f(x) has t and s with the same value 'r' so its not a function in the first place right??

    Please help a fellow understand. Pictures would be great.
    Hi Aquameatwad,

    g\circ f (x)= g(f (x))

    So domain of the function g\circ f is, \left\{t, s, k\right\}.

    If you take the element t, it maps to r under f, so;

    g\circ f (t)= g(f (t)) = g(r)

    But the element r is not included in the domain of g; hence, g(r) is not defined. Similar situation occurs for both s and k as well. Hence the set of ordered pairs of g\circ f is empty.

    i.e: g\circ f (x)= g(f (x)) = \emptyset
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  3. #3
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    Re: Constructions of functions

    I'm still confused. By using that idea wouldn't fg be empty as well? there is no 'l' or 'r' in g(x) either.
    And i don't know what you mean by "maps."
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  4. #4
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    Re: Constructions of functions

    let's look at f and g each as a series of "links". so when we say f(k) = l, we mean f takes k as a source, and links it to a target, l.

    here's f:

    k-->l
    s-->r
    t-->r the fact that t and k both go to r, does not violate the definition of a function, for example the squaring function takes both -2 and 2 to 4. if k went to r and l, THAT would be a problem.

    now, here's g:

    k-->s
    t-->s
    s-->k

    now fog (or "f follows g") means we start with possible starting places for g. if any of those end up at a starting place for f, we can continue:

    ..g.....f....

    k-->s-->r (g takes k to s, f takes s to r).
    t-->s-->r (g takes t to s, f takes s to r).
    s-->k-->l (g takes s to k, f takes k to l).

    this means fog = {(k,r), (t,r), (s,l)}

    what happens when we try to define gof ("g follows f")?

    ...f....g....

    k-->l-->?
    s-->r-->?
    t-->r-->? neither l nor r is a "starting point" for g, so we don't have any target defined for gof(k), gof(s) or gof(t).
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  5. #5
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    Re: Constructions of functions

    Thank you deveno i totally understand now. I drew a picture with your representations and its makes sense.
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