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.

Question

anonymous · Answer

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

swissgirl · Answer

I was wrongggg lol

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

anonymous · Answer

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 · Answer

now we need to solve 
$$0<-x^2+8x-4<4$$

anonymous · Answer

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 · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

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

anonymous · Answer

yw