Let f(x)=1/(x^2+1) and g(x)=x^-6. Find f(g(x)) and its domain.

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Let f(x)=1/(x^2+1) and g(x)=x^-6. Find f(g(x)) and its domain.

Mathematics
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I got 1/((x^-6)^2 + 1) which simplifies to x^3/(1+x^3)
That makes the domain all reals except for -1 the way I am looking at it, but that domain is not an option given.

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The options for the domain are all real numbers x>0 x is not equal to 0 x is greater than or equal to 0
Is it all real numbers since the powers were originally even?
I thought the domain had to be real at each step though.
\[f(x)=\frac{1}{x^2+1} \text{ and we have } g(x)=x^{-6} \text{ aka } g(x)=\frac{1}{x^6}\]
Right.
so notice the domains of f and g separately first
the domain of f is all real numbers but what is the domain of g? -- and also I'm having trouble seeing how you got your f(g(x)) anyways
Domain of g is anything but 0
|dw:1441048089386:dw|
so we already know we can't include 0 in the domain of f(g(x))
right so you should have |dw:1441134566849:dw|
and you can get rid of the negative exponents by multiplying top and bottom by x^(12)
Oh, duh! I don't know what I was thinking. Thank you!
My simplification is: \[\large \frac{1}{(x^{-6})^{2}+1}=\frac{x^{12}}{1+x^{12}}\]
you do understand the domain is all real numbers but x=0 right?
Yeah, I got it now. Thanks so much, both of you! :-)
this is because g(x) did not exist at x=0 you originally have \[f(g(x))=\frac{1}{(\frac{1}{x^6})^{2}+1} \\ \text{ and pluggin \in 0 into this makes that } \\ \text{ one little fraction undefined }\]

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