$\displaystyle f(x) = \displaystyle\frac{x}{1+x}$
$\displaystyle g(x) = sin2x$
Find
$\displaystyle (fog, gof,fof,gog)$
Really I'm wondering if these can be simplified after putting each expression into the proper function.
$\displaystyle f(x) = \displaystyle\frac{x}{1+x}$
$\displaystyle g(x) = sin2x$
Find
$\displaystyle (fog, gof,fof,gog)$
Really I'm wondering if these can be simplified after putting each expression into the proper function.
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}.
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.
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.
The domain is "sin(2x) not = -1" what values of x satisfy that?Domain {x|x not= -1)
$\displaystyle fof(x)= f(f(x))= f(\frac{x}{1+x})= \frac{\frac{x}{1+x}}{1+ \frac{1}{1+x}}$$\displaystyle fof = \displaystyle\frac{x}{2+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!
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))!Domain {x|x not=-1,-2}
Yes, that is correct. Your only problem was fof and since you got the other correct, I assume that was just a careless errror.$\displaystyle gog = sin2(sin2x)$
Domain = All real numbers