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for \(f(g(x))\) you must make sure that \(1
I was wrongggg lol
since \(f\) is only defined for numbers between 1 and 6 solve via \[1<5-x<6\] \[-4<-x<1\] \[4>x>-1\] or if you prefer \[-1
Yes. I got for that: \[Domain~of~f(g(x) = -1 ≤ x ≤ 4\]I seem to get irrational answers for the other part though as in \(g(f(x))\)
\(f(g(3))\) is defined because \(-1<3<4\)
Oh. Could I just plug in 3 here: \[g(f(x)) = 0 ≤ -x^2 + 8x - 4 ≤ 7\]and see if it is true or not then?
now we need to solve \[0<-x^2+8x-4<4\]
well that would answer the question for sure, but it would not find the domain of \(g\circ f\) it would only show that 3 is not in it
i think you mean find \(f(3)\) and see if that number is between 0 and 7
Well I now know how to find the domain of it if I wanted to, but this is a no calculator question, so it'd be hard to gauge the domain. I think I get it now because this is not true, it is undefined. \[0 ≤ 11 ≤ 7\]
yes that would make it undefined if you want to solve \[0<-x^2+8x-4<4\] you need to solve two separate inequalities, \[0<-x^2+8x-4\]and \[-x^2+8x-4<7\] and take the intersection
i keep putting 4 where there should be a 7, sorry
Oh ok. Thank you :) I get it now!