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Composition of Functions

  1. Aug 26, 2009 #1
    I have a homework problem as follows:

    Let f(x) = 2x, (-inf < x < inf). Can you think of functions g and h which satisfy the two equations g \circ f = 2gh and h \circ f = h^2 - g^2?

    I know that g \circ f = g(f(x)), which then = g(2x) for the first question. I can't figure out how to work the h in there.

    Help, please! Thank you.
     
  2. jcsd
  3. Aug 26, 2009 #2

    Dick

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    So you have g(2x)=2*g(x)*h(x) and h(2x)=h(x)^2-g(x)^2, right? I really can't think of any way to actually solve for g(x) and h(x). You should probably treat it as a guessing game. Have you ever seen formulas like that before? Hint: think of trig functions.
     
  4. Aug 27, 2009 #3
    The problem I am having is more basic than that. I can't see how to work the h into the first equation. In other words, if I have g composed with f, how does the function h enter into it at all? Thank you.
     
  5. Aug 27, 2009 #4

    Dick

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    You don't have to 'work it in' at all. You have g(f(x))=2*g(x)*h(x), right? Since f(x)=2x that's just g(2x)=g(x)*h(x). That's your first equation. There's not much else to do with it.
     
  6. Aug 28, 2009 #5
    Pay attention to what dick said, he pretty much gave you the answer.
    If you look at a table of trig identities, you will happen to come across 2 that look extremely similar to your problem.

    Don't feel bad, I had this same about a week ago and it took me like 5 days to figure it out. And i was so mad because it was this simple. Oh well, that's how math is.








    [physics/math double major OLE MISS] HOTTY TODDY!!!!!!!!
     
    Last edited: Aug 28, 2009
  7. Aug 31, 2009 #6
    Thank you both! Duh. The trig identities did it.
     
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