if f(g(x)=2x/(x+z), and g(x)=2x/(x+4)
find f(x)

- dtan5457

if f(g(x)=2x/(x+z), and g(x)=2x/(x+4)
find f(x)

- chestercat

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- dtan5457

- jim_thompson5910

is the `x+z` supposed to be `x+4` ?

- dtan5457

they are both under 2x so no

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## More answers

- jim_thompson5910

then I'm not sure. That z is throwing me off

- dtan5457

perhaps its an x+2? I mean my teacher's handwriting does look like a Z.

- jim_thompson5910

that sounds possible

- anonymous

i have an answer that might work, but i think you are supposed to use inverse functions some how

- anonymous

by trial and error i came up with \[f(x)=\frac{1}{x-\frac{2}{x}}\]] we can see if that works, and also clean it up a bit

- anonymous

in any case it is wrong, but i think i can fix it

- anonymous

lets try this one \[f(x)=\frac{1}{\frac{1}{x}-\frac{2}{2x}}\]

- anonymous

ok one more time without me being so stupid \[f(x)=\frac{1}{\frac{1}{x}-\frac{2}{x+4}}\]

- Nnesha

*

- anonymous

i think that actually works

- dtan5457

where would a Z come up though? or is that for x+2

- anonymous

i used 2
with z it is too weird

- dtan5457

so it would be impossible with z?

- anonymous

i am wagering it is 2

- anonymous

i wonder if we can do it with differentiation though, is that part of this class?

- dtan5457

this is pre calc we only did difference quotients so far

- anonymous

maybe @Zarkon has a better idea
seems like it needs to use inverse functions some how but i am clogged up
in any case i am almost certain the (third) answer i wrote above works, just trying to figure how to subtract 2 in the denominator

- anonymous

take the reciprocal, subtract 2 in the numerator, then take the reciprocal again
of course that mess i wrote can be cleaned up without using compound fractions

- dtan5457

my teacher may have wrote something wrong of some sort its only been like 2 weeks into school i doubt he would have something harder than you can solve

- anonymous

usually you are told \(f\) and \(f\circ g\) and asked to find \(g\)
that you can do by finding the inverse of \(f\) and applying it
this is a bit different

- dtan5457

you are right in my class example we had to find g and that was fairly simple

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