Question

if f(x)= (x+1)/(x^2-1) and g(x) = (3x+7)/2x
find f[g(x)].

Please answer clearly and explain the solution clearly and easy to understand. Thanks!
(guaranteed medal for a good answer)

Answer 1

step on write
\[f(g(x))\]

Answer 2

step two replace
\[g(x)\] by
\[\frac{3x+7}{2x}\] to get
\[f(\frac{3x+7}{2x})\]

Answer 3

so far so good?

Answer 4

yup

Answer 5

then were we see an x in \[f(x)\] we replace it by
\[\frac{3x+7}{2x}\] to get
\[\frac{\frac{3x+7}{2x}+1}{(\frac{3x+7}{2x})^2-1}\]

Answer 6

zat ok?

Answer 7

is jsut an exercise in simplication

Answer 8

simplification

Answer 9

well actually why don't we simplify to begin with.
\[f(x)=\frac{x+1}{x^2-1}=\frac{x+1}{(x+1)(x-1)}=\frac{1}{x-1}\]\]

Answer 10

yup that is what my teacher told me

Answer 11

so we could make this problem much easier if we use the second f i wrote. on the other hand this is not really valid

Answer 12

those function are not identical. they are identical if x is not -1. if x is -1 the first function is undefined, whereas the second function is
\[\frac{1}{-2}\]

Answer 13

but if your teacher does not mind, neither do i. just write
\[f(x)=\frac{1}{x-1}\] and so \[f(g(x))=f(\frac{3x+7}{2x})=\frac{1}{\frac{3x+7}{2x}-1}\]

Answer 14

what??

Answer 15

then multiply top and bottom by 2x to get
\[f(g(x))=\frac{2x}{3x+7-2x}=\frac{2x}{x+7}\]\][

Answer 16

what happened to the ( )^2?

Answer 17

oops i lost you

Answer 18

are we going to replace \[f(x)=\frac{x+1}{x^2-1}\] by
\[f(x)=\frac{1}{x-1}\]?

Answer 19

yes because i think it is easier

Answer 20

ok in that case there is no square in the denominator

Answer 21

ooooohhh now i know

Answer 22

if we make that change in f, then \[f(g(x))=f(\frac{3x+7}{2x})\] yes?

Answer 23

if so then everything i wrote above works yes?

Answer 24

if not let me know

Answer 25

isn't it it should first be 1/(3x+7)- 1?

Answer 26

am i correct?

Answer 27

no.

Answer 28

why?

Answer 29

\[g(x)=\frac{3x+7}{2x}\]

Answer 30

not
\[3x+7\]

Answer 31

so when we replace x in f(x) by \[\frac{3x+7}{2x}\] we get \[\frac{1}{\frac{3x+7}{2x}-1}\]

Answer 32

yup i forgot to type 2x sorry for that

Answer 33

please continue

Answer 34

ok good

Answer 35

now you can simplify this compound fraction by multiplying top and bottom by 2x

Answer 36

we have
\[f(g(x))=\frac{1}{\frac{3x+7}{2x}-1}\]
\[=\frac{1}{\frac{3x+7}{2x}-1}\times \frac{2x}{2x}\]
\[=\frac{2x}{3x+7-2x}\]
\[=\frac{2x}{x+7}\]

Answer 37

k?

Answer 38

what happened with the 2x below 3x+7?

Answer 39

i multiplied top and bottom by 2x to get rid of it. canceled from the denominator

Answer 40

\[\frac{3x+7}{2x}\times 2x=3x+7\]

Answer 41

ohhh ok

Answer 42

you cannot have a compound fraction as an answer, so i got rid of it by multiplying top and bottom by that annoying 2x in the denominator of the denominator

Answer 43

clear or no?

Answer 44

wait ill do it on a paper ^_^

Answer 45

k

Answer 46

how did it become 2x/2x+7?

Answer 47

ok let us start with
\[\frac{1}{\frac{3x+7}{2x}-1}\] ok?

Answer 48

ok

Answer 49

multiply top and bottom by 2x. what do you get?

Answer 50

\[1\times 2x=2x\] that is your numerator

Answer 51

ok

Answer 52

\[\frac{3x+7}{2x}\times 2x = 3x+7\] that is why you did it to get rid of the pesky 2x in the bottom

Answer 53

\[-1\times 2x=-2x\]

Answer 54

ok

Answer 55

of course you have to multiply the whole denominator by 2x, not just part of it.

Answer 56

so you numerator is 2x

Answer 57

your denominator is 3x+7-2x

Answer 58

which give a denominator of x+7

Answer 59

ok

Answer 60

so our "final answer" is \[f(g(x))=\frac{2x}{x+7}\]

Answer 61

i just got confused i forgot that 3x and -2x can be combined

Answer 62

ah but clear now yes?

Answer 63

yes! thanks^_^

Answer 64

welcome

Answer 65

ah but clear now yes?