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Composition of Functions

  • #1
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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.
 

Answers and Replies

  • #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.
 
  • #3
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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.
 
  • #4
Dick
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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.
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.
 
  • #5
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.
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:
  • #6
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Thank you both! Duh. The trig identities did it.
 

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