Question

For f(x)=(1-x)/(1+x) and g(x) = (x)/(1-x) find the simplified form for f [g(x)] and state the domain.

*this is an essay question so im going to need to understand every step of the problem and explain how i got the solution*** :)

@ganeshie8

Answer 1

show what you have so far

Answer 2

f(g(x)) means you plug in g(x) where ever there is an x in f(x) basically

Answer 3

As this is a essay question, you should show all your steps as ganeshie required you to and we'll just point out if you're on the right path or not, because if we do this problem for you, then you'll learn nothing.

Answer 4

okay so I guess i can say that I know: I take g(x) and plug it into f(X) correct?

Answer 5

yes that's what you just said, so , is that what they mean by the simplified form? also what does it mean by saying: state the domain? :)

Answer 6

also im not sure what the next step is after plugging in g(x) into where i see x in f(X).
This is what i got:
f(x) = (1- (x/1-x)) / (1+(x/1-x))

Answer 7

The domain is the set of values for which the function is defined, so the function has to give you real numbers.
For your case you'll be looking at where it divides by 0, and taking the square root of negative numbers that's where it will be restricted

Answer 8

so okay im a little shaky on that portion, but what comes after this: f(x) = 1 - (x/1-x) / 1+(x/1-x)

Answer 9

that simplifies further..

Answer 10

yes how do i do that

Answer 11

? :)

Answer 12

Well lets see what you have, I need to put it in latex haha,
\[\huge f(g(x)) = \frac{ (1-\frac{ x }{ (1-x) } )}{ 1+\frac{ x }{ (1-x) } }\]

Answer 13

ok yes what happens next ? :)

Answer 14

i have to make the bottoms the same to subtract, so how do i do that step? :)

Answer 15

There's a few ways to do this I guess, you can find a common denominator, that maybe easy but tedious >.<

Answer 16

or the one in which u recommend

Answer 17

Prob tedious anyway, and idk how good you are with algebra, lets just find common denominator for the numerator and denominator if that made any sense at all

Answer 18

\[1-\frac{ x }{ (1-x) }\] so do this first and then for the denominator as well

Answer 19

okay so do i multiply both sides with (1-x)

Answer 20

\[\frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]

Answer 21

yes :)

Answer 22

Yeah, keep working on it, I think you're on the right track ;)

Answer 23

okay :)
f[g(x)]= f(x/1-x) = 1 - (x/1-x) / 1+(x/1-x) = [(1-x-x)/(1-x)] / [(1-x+x)/(1-x)]

Answer 24

simplifies further..

Answer 25

^ ahahhh

Answer 26

Keep working at it math

Answer 27

ok :)
[(1-2x/(1-x)]/[(1/1-x)]

Answer 28

Then i'll divide so i'll get:
-2x+1/1-x * 1-x/1 and the 1-x will cross out so i can get -2x+1 as a final answer

Answer 29

That worked out pretty well

Answer 30

is it correct ? :D now i have to find the domainnnn :P

Answer 31

Looking at the original function, what can we say would make it undefined

Answer 32

the 0 in the denominator or taking the sqrt of - numbers

Answer 33

so it is x is greater than or equal to 0 because you cannot have a negative number inside a sqrt?

Answer 34

actually no we dont have a sqrt lol

Answer 35

so how does it divide by 0?

Answer 36

No we don't so I mean like if we have this for example

Answer 37

\[f(x) = \frac{ x }{ x^2-4 }\] what values of x would make this undefined

Answer 38

so its x is greater than 0
:)?

Answer 39

2 & -2

Answer 40

\[x^2-4 \neq 0\]

Answer 41

Yeah

Answer 42

So yours would be x cannot = 1 right?

Answer 43

OOHH x cannot equal -1!!

Answer 44

yess -1&1 because for one equation there is a minus sign & for another there is a plus.

Answer 45

-1 is not a bad value for the domain

Answer 46

Domain of f(g(x)) should be \[x\ne 1\]

Answer 47

In words : all real values except 1

Answer 48

you just need to look at the domain of inner function g(x) and the final simplified form

Answer 49

well if i look @ the final simpliifed form, I wouldn't have a problem with the domain because my denominator is 1. correct?

Answer 50

@ganeshie8

Answer 51

@ikram002p i have a question, can u explain to me why the domain here is : all reals except 1?

Answer 52

@Destinymasha @Destinymasha

Answer 53

domain of g(x) sice if its 1 u would have 1/0

Answer 54

why do you look @ the domain of g(x) and not the domain of f(x)

Answer 55

@ikram002p

Answer 56

domain of f(x) is every where except x=-1 not 1 :)

Answer 57

f(g(x))

to evaluate this all it matters are g(x) and f(g(x))
you're not touching f(x) directly..