anonymous
  • anonymous
Let \(f\) and g be defined by\[f(x) = -x^2 + 8x - 4~~~~~1 ≤ x ≤ 6\]\[g(x) = 5 - x~~~~~0 ≤ x ≤ 7\]How would I go about with finding the domains of \(f(g(x))\) and \(g(f(x))\)? Also, how would I prove that \(f(g(3))\) is defined and \(g(f(3))\) is undefined? More than likely, it has to do with the domain, but I'm not fully sure with how to do it.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
for \(f(g(x))\) you must make sure that \(1
swissgirl
  • swissgirl
I was wrongggg lol
anonymous
  • anonymous
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

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anonymous
  • anonymous
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 \(g(f(x))\)
anonymous
  • anonymous
\(f(g(3))\) is defined because \(-1<3<4\)
anonymous
  • anonymous
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?
anonymous
  • anonymous
now we need to solve \[0<-x^2+8x-4<4\]
anonymous
  • anonymous
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
anonymous
  • anonymous
i think you mean find \(f(3)\) and see if that number is between 0 and 7
anonymous
  • anonymous
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. \[0 ≤ 11 ≤ 7\]
anonymous
  • anonymous
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
anonymous
  • anonymous
i keep putting 4 where there should be a 7, sorry
anonymous
  • anonymous
Oh ok. Thank you :) I get it now!
anonymous
  • anonymous
yw

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