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Calcmathlete

  • 2 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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  1. satellite73
    • 2 years ago
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    for \(f(g(x))\) you must make sure that \(1<g(x)<6\)

  2. swissgirl
    • 2 years ago
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    I was wrongggg lol

  3. satellite73
    • 2 years ago
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    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 \[-1<x<4\]

  4. Calcmathlete
    • 2 years ago
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    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 \(g(f(x))\)

  5. satellite73
    • 2 years ago
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    \(f(g(3))\) is defined because \(-1<3<4\)

  6. Calcmathlete
    • 2 years ago
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    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?

  7. satellite73
    • 2 years ago
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    now we need to solve \[0<-x^2+8x-4<4\]

  8. satellite73
    • 2 years ago
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    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

  9. satellite73
    • 2 years ago
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    i think you mean find \(f(3)\) and see if that number is between 0 and 7

  10. Calcmathlete
    • 2 years ago
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    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. \[0 ≤ 11 ≤ 7\]

  11. satellite73
    • 2 years ago
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    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

  12. satellite73
    • 2 years ago
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    i keep putting 4 where there should be a 7, sorry

  13. Calcmathlete
    • 2 years ago
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    Oh ok. Thank you :) I get it now!

  14. satellite73
    • 2 years ago
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    yw

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