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Thread: Functions with sin2x

  1. #1
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    Functions with sin2x

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

    $\displaystyle g(x) = sin2x$

    Find
    $\displaystyle (fog, gof,fof,gog)$

    Really I'm wondering if these can be simplified after putting each expression into the proper function.
    Last edited by Latvija13; Jan 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
    $\displaystyle f(x) = x/1+x$
    $\displaystyle g(x) = sin2x$
    Find
    $\displaystyle (fog, gof,fof,gog)$

    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 $\displaystyle \sin(2x) \ \ \text{or} \ \ 2\sin(x)\cos(x)$
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    I tried $\displaystyle 2sin(x)cos(x)$ in the first one, fog, and came up with

    $\displaystyle 2sin(x)cos(x)/1+2sin(x)cos(x)$

    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 $\displaystyle 2sin(x)cos(x)$ in the first one, fog, and came up with

    $\displaystyle 2sin(x)cos(x)/1+2sin(x)cos(x)$

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

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

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

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

    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.

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

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

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

    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:

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

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

    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:

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

    appears to work:
    $\displaystyle sin\displaystyle\frac{2x}{1+x}$
    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
    $\displaystyle f(x) = \displaystyle\frac{x}{1+x}$

    $\displaystyle g(x) = sin2x$

    Find
    $\displaystyle (fog, gof,fof,gog)$

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

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

    Domain {x|x not= -1)

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

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

    $\displaystyle gog = sin2(sin2x)$

    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:

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

    Domain {x|x not= -1)

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

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

    $\displaystyle gog = sin2(sin2x)$

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

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

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

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

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

    $\displaystyle gog = sin2(sin2x)$

    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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