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anonymous
 4 years ago
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.
anonymous
 4 years ago
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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for \(f(g(x))\) you must make sure that \(1<g(x)<6\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since \(f\) is only defined for numbers between 1 and 6 solve via \[1<5x<6\] \[4<x<1\] \[4>x>1\] or if you prefer \[1<x<4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. 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
 4 years ago
Best ResponseYou've already chosen the best response.0\(f(g(3))\) is defined because \(1<3<4\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh. 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
 4 years ago
Best ResponseYou've already chosen the best response.0now we need to solve \[0<x^2+8x4<4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well 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
 4 years ago
Best ResponseYou've already chosen the best response.0i think you mean find \(f(3)\) and see if that number is between 0 and 7

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well 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
 4 years ago
Best ResponseYou've already chosen the best response.0yes that would make it undefined if you want to solve \[0<x^2+8x4<4\] you need to solve two separate inequalities, \[0<x^2+8x4\]and \[x^2+8x4<7\] and take the intersection

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i keep putting 4 where there should be a 7, sorry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh ok. Thank you :) I get it now!
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