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!
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- moongazer

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- anonymous

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

- anonymous

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

- anonymous

so far so good?

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## More answers

- moongazer

yup

- anonymous

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

- anonymous

zat ok?

- anonymous

is jsut an exercise in simplication

- anonymous

simplification

- anonymous

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

- moongazer

yup that is what my teacher told me

- anonymous

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

- anonymous

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
\[\frac{1}{-2}\]

- anonymous

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

- moongazer

what??

- anonymous

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

- moongazer

what happened to the ( )^2?

- anonymous

oops i lost you

- anonymous

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

- moongazer

yes because i think it is easier

- anonymous

ok in that case there is no square in the denominator

- moongazer

ooooohhh now i know

- anonymous

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

- anonymous

if so then everything i wrote above works yes?

- anonymous

if not let me know

- moongazer

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

- moongazer

am i correct?

- anonymous

no.

- moongazer

why?

- anonymous

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

- anonymous

not
\[3x+7\]

- anonymous

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

- moongazer

yup i forgot to type 2x sorry for that

- moongazer

please continue

- anonymous

ok good

- anonymous

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

- anonymous

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

- anonymous

k?

- moongazer

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

- anonymous

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

- anonymous

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

- moongazer

ohhh ok

- anonymous

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

- anonymous

clear or no?

- moongazer

wait ill do it on a paper ^_^

- anonymous

k

- moongazer

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

- anonymous

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

- moongazer

ok

- anonymous

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

- anonymous

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

- moongazer

ok

- anonymous

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

- anonymous

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

- moongazer

ok

- anonymous

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

- anonymous

so you numerator is 2x

- anonymous

your denominator is 3x+7-2x

- anonymous

which give a denominator of x+7

- moongazer

ok

- anonymous

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

- moongazer

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

- anonymous

ah but clear now yes?

- moongazer

yes! thanks^_^

- anonymous

welcome

- anonymous

ah but clear now yes?

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