## Calcmathlete 3 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.

1. satellite73

for \(f(g(x))\) you must make sure that \(1<g(x)<6\)

2. swissgirl

I was wrongggg lol

3. satellite73

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

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

\(f(g(3))\) is defined because \(-1<3<4\)

6. Calcmathlete

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

now we need to solve \[0<-x^2+8x-4<4\]

8. satellite73

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

i think you mean find \(f(3)\) and see if that number is between 0 and 7

10. Calcmathlete

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

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

i keep putting 4 where there should be a 7, sorry

13. Calcmathlete

Oh ok. Thank you :) I get it now!

14. satellite73

yw