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Natural domain of
\[f(x)=\frac{ 1 }{ x-3 }\]
g(x)=5x
\[(Fog)(x)\]
It is x\[x \in[-1,infinite)\] ?

what did you get for f(g(x)) = .... ?

Let me do it, give me a sec

sure, take your time :)

\[\sqrt{2-4x}\]

ooops sorry, confused question, give me another sec

am i reading the question correct ?
f(x) =1/ (x-3)
g(x) =5x
?
i don't see .....
oh, okk

Sorry, the mistake was the question

\[f(x)=\sqrt{x+1}\]
g(x)=1-4x

yes, f(g(x)) is correct then

now quantity under square root cannot be negative

So the domain is x<= 1/2 for the square root, and I take the intersection

[-1,1/2]

yep, thats correct! good :)

So I need to take the domain of FoG with
\[\left\{ x \in Dg|g(x)\in Df \right\}\]

And intersect that with the domain of the last result?

Perfect, thanks

welcome ^_^