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)

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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)

Mathematics
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step on write \[f(g(x))\]
step two replace \[g(x)\] by \[\frac{3x+7}{2x}\] to get \[f(\frac{3x+7}{2x})\]
so far so good?

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Other answers:

yup
then were we see an x in \[f(x)\] we replace it by \[\frac{3x+7}{2x}\] to get \[\frac{\frac{3x+7}{2x}+1}{(\frac{3x+7}{2x})^2-1}\]
zat ok?
is jsut an exercise in simplication
simplification
well actually why don't we simplify to begin with. \[f(x)=\frac{x+1}{x^2-1}=\frac{x+1}{(x+1)(x-1)}=\frac{1}{x-1}\]\]
yup that is what my teacher told me
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
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 \[\frac{1}{-2}\]
but if your teacher does not mind, neither do i. just write \[f(x)=\frac{1}{x-1}\] and so \[f(g(x))=f(\frac{3x+7}{2x})=\frac{1}{\frac{3x+7}{2x}-1}\]
what??
then multiply top and bottom by 2x to get \[f(g(x))=\frac{2x}{3x+7-2x}=\frac{2x}{x+7}\]\][
what happened to the ( )^2?
oops i lost you
are we going to replace \[f(x)=\frac{x+1}{x^2-1}\] by \[f(x)=\frac{1}{x-1}\]?
yes because i think it is easier
ok in that case there is no square in the denominator
ooooohhh now i know
if we make that change in f, then \[f(g(x))=f(\frac{3x+7}{2x})\] yes?
if so then everything i wrote above works yes?
if not let me know
isn't it it should first be 1/(3x+7)- 1?
am i correct?
no.
why?
\[g(x)=\frac{3x+7}{2x}\]
not \[3x+7\]
so when we replace x in f(x) by \[\frac{3x+7}{2x}\] we get \[\frac{1}{\frac{3x+7}{2x}-1}\]
yup i forgot to type 2x sorry for that
please continue
ok good
now you can simplify this compound fraction by multiplying top and bottom by 2x
we have \[f(g(x))=\frac{1}{\frac{3x+7}{2x}-1}\] \[=\frac{1}{\frac{3x+7}{2x}-1}\times \frac{2x}{2x}\] \[=\frac{2x}{3x+7-2x}\] \[=\frac{2x}{x+7}\]
k?
what happened with the 2x below 3x+7?
i multiplied top and bottom by 2x to get rid of it. canceled from the denominator
\[\frac{3x+7}{2x}\times 2x=3x+7\]
ohhh ok
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
clear or no?
wait ill do it on a paper ^_^
k
how did it become 2x/2x+7?
ok let us start with \[\frac{1}{\frac{3x+7}{2x}-1}\] ok?
ok
multiply top and bottom by 2x. what do you get?
\[1\times 2x=2x\] that is your numerator
ok
\[\frac{3x+7}{2x}\times 2x = 3x+7\] that is why you did it to get rid of the pesky 2x in the bottom
\[-1\times 2x=-2x\]
ok
of course you have to multiply the whole denominator by 2x, not just part of it.
so you numerator is 2x
your denominator is 3x+7-2x
which give a denominator of x+7
ok
so our "final answer" is \[f(g(x))=\frac{2x}{x+7}\]
i just got confused i forgot that 3x and -2x can be combined
ah but clear now yes?
yes! thanks^_^
welcome
ah but clear now yes?

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