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Math Help - Functions with sin2x

  1. #1
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    Functions with sin2x

    f(x) = \displaystyle\frac{x}{1+x}

    g(x) = sin2x

    Find
    (fog, gof,fof,gog)

    Really I'm wondering if these can be simplified after putting each expression into the proper function.
    Last edited by Latvija13; January 26th 2011 at 03:26 PM. Reason: Needed to fix the expressions
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  2. #2
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    Quote Originally Posted by Latvija13 View Post
    f(x) = x/1+x
    g(x) = sin2x
    Find
    (fog, gof,fof,gog)

    Really I'm wondering if these can be simplified after putting each expression into the proper function.
    It depends on what you consider simplified. Would you rather have \sin(2x) \ \ \text{or} \ \ 2\sin(x)\cos(x)
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    I tried 2sin(x)cos(x) in the first one, fog, and came up with

    2sin(x)cos(x)/1+2sin(x)cos(x)

    I'm just not use to working with both trig and non trig functions.
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  4. #4
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    Quote Originally Posted by Latvija13 View Post
    I tried 2sin(x)cos(x) in the first one, fog, and came up with

    2sin(x)cos(x)/1+2sin(x)cos(x)

    I'm just not use to working with both trig and non trig functions.
    \displaystyle\frac{2\sin(x)\cos(x)}{1}=2\sin(x)\co  s(x)

    \displaystyle 2sin(x)cos(x)/1+2sin(x)cos(x)=4\sin(x)\cos(x)=2\sin(2x)
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\frac{2\sin(x)\cos(x)}{1}=2\sin(x)\co  s(x)

    \displaystyle 2sin(x)cos(x)/1+2sin(x)cos(x)=4\sin(x)\cos(x)=2\sin(2x)
    I'm sorry, it should have been.

    \displaystyle\frac{2\sin(x)\cos(x)}{1+2sin(x)cos(x  )}

    I didn't know how to use the fraction bar.
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  6. #6
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    Quote Originally Posted by Latvija13 View Post
    I'm sorry, it should have been.

    \displaystyle\frac{2\sin(x)\cos(x)}{1+2sin(x)cos(x  )}

    I didn't know how to use the fraction bar.
    either way ...

    \displaystyle\frac{2\sin(x)\cos(x)}{1+2sin(x)cos(x  )} = \frac{\sin(2x)}{1+\sin(2x)}

    simplicity is in the eye of the solver.
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  7. #7
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    And we're back where we started!

    The moral: not everything can be simplified!
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    How do I know which ones can be simplified? For instance, the next composite function g o f:

    sin2(\displaystyle\frac{x}{1+x})

    appears to work:
    sin\displaystyle\frac{2x}{1+x}

    Then I believe the domain would be {x|x not= 0}.
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  9. #9
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    Quote Originally Posted by Latvija13 View Post
    How do I know which ones can be simplified? For instance, the next composite function g o f:

    sin2(\displaystyle\frac{x}{1+x})

    appears to work:
    sin\displaystyle\frac{2x}{1+x}
    well, that's just multiplying the "2" in but, yes, it is a little "simplified".

    Then I believe the domain would be {x|x not= 0}.[/QUOTE]
    No. You cannot divide by 0 but here, the denominator is 1+ x. That is 0 when x= -1, not when x=0. If x= 0, that would be sin(0)= 0 which is well defined.
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    I think I'm starting to understand how these work. I was expecting them to work like identities and have multiple steps to get to the point where I can determine the domain, but it appears that isn't the true.
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  11. #11
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    Quote Originally Posted by Latvija13 View Post
    f(x) = \displaystyle\frac{x}{1+x}

    g(x) = sin2x

    Find
    (fog, gof,fof,gog)

    Really I'm wondering if these can be simplified after putting each expression into the proper function.
    Here is what I ended up with:

    fog=sin(2x)\displaystyle(\frac{x}{1+x})

    Domain {x|x not= -1)

    fof = \displaystyle\frac{x}{2+x}

    Domain {x|x not=-1,-2}

    gog = sin2(sin2x)

    Domain = All real numbers
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  12. #12
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    Quote Originally Posted by Latvija13 View Post
    Here is what I ended up with:

    fog=sin(2x)\displaystyle(\frac{x}{1+x})

    Domain {x|x not= -1)

    fof = \displaystyle\frac{x}{2+x}

    Domain {x|x not=-1,-2}

    gog = sin2(sin2x)

    Domain = All real numbers
    \displaystyle f\circ g = f(g(x))=\sin\left(\frac{2x}{x+1}\right)

    \displaystyle \text{dom}(f\circ f)=\{x|x\neq -2\}
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  13. #13
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    Quote Originally Posted by Latvija13 View Post
    Here is what I ended up with:

    fog=sin(2x)\displaystyle(\frac{x}{1+x})
    No. This is the product, fg(x). The composition, fog(x)= f(g(x))= f(sin(2x))= \frac{sin(2x)}{1+ sin(2x)} as dwsmith says.

    Domain {x|x not= -1)
    The domain is "sin(2x) not = -1" what values of x satisfy that?

    fof = \displaystyle\frac{x}{2+x}
     fof(x)= f(f(x))= f(\frac{x}{1+x})= \frac{\frac{x}{1+x}}{1+ \frac{1}{1+x}}
    Multipying both numerator and denominator by 1+ x makes that fof(x)= \frac{1}{1+x+ 1}= \frac{x}{2+ x} just as you have!

    Domain {x|x not=-1,-2}
    Very good! Although the formula you give is "calculable" at x= -1, f is not and you must be able to do that for f(f(-1))!

    gog = sin2(sin2x)

    Domain = All real numbers
    Yes, that is correct. Your only problem was fof and since you got the other correct, I assume that was just a careless errror.
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