Question

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

find f(x)

Answer 1

@jim_thompson5910

Answer 2

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

Answer 3

they are both under 2x so no

Answer 4

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

Answer 5

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

Answer 6

that sounds possible

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

*

Answer 13

i think that actually works

Answer 14

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

Answer 15

i used 2
with z it is too weird

Answer 16

so it would be impossible with z?

Answer 17

i am wagering it is 2

Answer 18

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

Answer 19

this is pre calc we only did difference quotients so far

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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