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How to Do (f o g)(1) Function Problem?

  1. Nov 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Use the following functions to answer questions:
    (f)(x) = 2x + 3
    g(x) = x + 4

    FIND (f o g)(1)


    2. Relevant equations

    Everything needed is above.

    3. The attempt at a solution

    I know that (f o g)(1) is a "composure" function problem and (f o g)(x) is the same as saying f(g(x)) - we've solved these before. However, in this problem, instead of an "x" we have a "1" inside "()."

    My question is what do we do in this case? For example, do we still create a composure problem? If so, how? Would it be something like:

    f(g(1))

    If so, what do we do? Just that "1" instead of an "x" is throwing me off here in what to do. Thanks very much for your help!
     
  2. jcsd
  3. Nov 20, 2014 #2
    Yes you're right (f o g)(1) is composition of functions, so f(g(1). So if f(x) = 2x + 3 then f(g(x)) = 2*g(x) + 3. Now that you have that, it should be pretty easy to find the answer.
     
  4. Nov 21, 2014 #3
    actually you have two equivalent ways to answer this problem ,
    The first one is to find g(1) then substitute the value pf g(1) in any x in the f(x)
    The other way , as you and @Panphobia said , is to do it like : f(x o g) (1) = 2g(1)+3
    .
    .
    They are equivalent , you will get the same answer .. (:
     
  5. Nov 22, 2014 #4

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    The easiest and most straightforward way to find f(g(1)) is to find g(1) first, and then plug that into f. When you do it this way, you don't even have to figure out what function ##f\circ g## is. The alternative is to figure out what function ##f\circ g## is first, and then plug 1 into ##f\circ g##.

    ##f\circ g## is defined by ##(f\circ g)(x)=f(g(x))## for all ##x## in the domain of g. The "for all" is essential. It means that the equality ##(f\circ g)(x)=f(g(x))## holds regardless of what number the symbol ##x## represents. So in particular, it holds when ##x=1##.
    This notation doesn't make sense.
     
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