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osansevieroBest ResponseYou've already chosen the best response.1
Natural domain of \[f(x)=\frac{ 1 }{ x3 }\] g(x)=5x \[(Fog)(x)\] It is x\[x \in[1,infinite)\] ?
 5 months ago

hartnnBest ResponseYou've already chosen the best response.1
what did you get for f(g(x)) = .... ?
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.1
Let me do it, give me a sec
 5 months ago

hartnnBest ResponseYou've already chosen the best response.1
sure, take your time :)
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.1
ooops sorry, confused question, give me another sec
 5 months ago

hartnnBest ResponseYou've already chosen the best response.1
am i reading the question correct ? f(x) =1/ (x3) g(x) =5x ? i don't see ..... oh, okk
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.1
Sorry, the mistake was the question
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.1
\[f(x)=\sqrt{x+1}\] g(x)=14x
 5 months ago

hartnnBest ResponseYou've already chosen the best response.1
yes, f(g(x)) is correct then
 5 months ago

hartnnBest ResponseYou've already chosen the best response.1
now quantity under square root cannot be negative
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.1
So the domain is x<= 1/2 for the square root, and I take the intersection
 5 months ago

hartnnBest ResponseYou've already chosen the best response.1
yep, thats correct! good :)
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.1
So I need to take the domain of FoG with \[\left\{ x \in Dgg(x)\in Df \right\}\]
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.1
And intersect that with the domain of the last result?
 5 months ago

hartnnBest ResponseYou've already chosen the best response.1
i am not sure about what that notation says, but yes, you take the intersection domains of both the funtion with composite function to get the final domain
 5 months ago
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