Functions with sin2x

**Latvija13** · January 26th 2011, 02:48 PM

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

$g(x) = sin2x$

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

Really I'm wondering if these can be simplified after putting each expression into the proper function.

**dwsmith** · January 26th 2011, 02:56 PM

Originally Posted by Latvija13

$f(x) = x/1+x$
$g(x) = sin2x$
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 $\sin(2x) \ \ \text{or} \ \ 2\sin(x)\cos(x)$

**Latvija13** · January 26th 2011, 03:04 PM

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.

**dwsmith** · January 26th 2011, 03:07 PM

Originally Posted by Latvija13

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.

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

$\displaystyle 2sin(x)cos(x)/1+2sin(x)cos(x)=4\sin(x)\cos(x)=2\sin(2x)$

**Latvija13** · January 26th 2011, 03:15 PM

Originally Posted by dwsmith

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

$\displaystyle 2sin(x)cos(x)/1+2sin(x)cos(x)=4\sin(x)\cos(x)=2\sin(2x)$

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.

**skeeter** · January 26th 2011, 04:17 PM

Originally Posted by Latvija13

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.

**HallsofIvy** · January 26th 2011, 04:39 PM

And we're back where we started!

The moral: not everything can be simplified!

**Latvija13** · January 26th 2011, 04:53 PM

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}.

**HallsofIvy** · January 26th 2011, 04:58 PM

Originally Posted by Latvija13

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.

**Latvija13** · January 26th 2011, 05:18 PM

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.

**Latvija13** · January 26th 2011, 07:04 PM

Originally Posted by Latvija13

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

$g(x) = sin2x$

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

**dwsmith** · January 26th 2011, 07:29 PM

Originally Posted by Latvija13

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

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

$\displaystyle \text{dom}(f\circ f)=\{x|x\neq -2\}$

**HallsofIvy** · January 27th 2011, 05:07 AM

Originally Posted by Latvija13

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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