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moongazer
 5 years ago
if f(x)= (x+1)/(x^21) 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)
moongazer
 5 years ago
if f(x)= (x+1)/(x^21) 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0step on write \[f(g(x))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0step two replace \[g(x)\] by \[\frac{3x+7}{2x}\] to get \[f(\frac{3x+7}{2x})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then were we see an x in \[f(x)\] we replace it by \[\frac{3x+7}{2x}\] to get \[\frac{\frac{3x+7}{2x}+1}{(\frac{3x+7}{2x})^21}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is jsut an exercise in simplication

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well actually why don't we simplify to begin with. \[f(x)=\frac{x+1}{x^21}=\frac{x+1}{(x+1)(x1)}=\frac{1}{x1}\]\]

moongazer
 5 years ago
Best ResponseYou've already chosen the best response.0yup that is what my teacher told me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0those 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
 5 years ago
Best ResponseYou've already chosen the best response.0but if your teacher does not mind, neither do i. just write \[f(x)=\frac{1}{x1}\] and so \[f(g(x))=f(\frac{3x+7}{2x})=\frac{1}{\frac{3x+7}{2x}1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then multiply top and bottom by 2x to get \[f(g(x))=\frac{2x}{3x+72x}=\frac{2x}{x+7}\]\][

moongazer
 5 years ago
Best ResponseYou've already chosen the best response.0what happened to the ( )^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are we going to replace \[f(x)=\frac{x+1}{x^21}\] by \[f(x)=\frac{1}{x1}\]?

moongazer
 5 years ago
Best ResponseYou've already chosen the best response.0yes because i think it is easier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok in that case there is no square in the denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if we make that change in f, then \[f(g(x))=f(\frac{3x+7}{2x})\] yes?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if so then everything i wrote above works yes?

moongazer
 5 years ago
Best ResponseYou've already chosen the best response.0isn't it it should first be 1/(3x+7) 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[g(x)=\frac{3x+7}{2x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so when we replace x in f(x) by \[\frac{3x+7}{2x}\] we get \[\frac{1}{\frac{3x+7}{2x}1}\]

moongazer
 5 years ago
Best ResponseYou've already chosen the best response.0yup i forgot to type 2x sorry for that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you can simplify this compound fraction by multiplying top and bottom by 2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have \[f(g(x))=\frac{1}{\frac{3x+7}{2x}1}\] \[=\frac{1}{\frac{3x+7}{2x}1}\times \frac{2x}{2x}\] \[=\frac{2x}{3x+72x}\] \[=\frac{2x}{x+7}\]

moongazer
 5 years ago
Best ResponseYou've already chosen the best response.0what happened with the 2x below 3x+7?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i multiplied top and bottom by 2x to get rid of it. canceled from the denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{3x+7}{2x}\times 2x=3x+7\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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

moongazer
 5 years ago
Best ResponseYou've already chosen the best response.0wait ill do it on a paper ^_^

moongazer
 5 years ago
Best ResponseYou've already chosen the best response.0how did it become 2x/2x+7?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok let us start with \[\frac{1}{\frac{3x+7}{2x}1}\] ok?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0multiply top and bottom by 2x. what do you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1\times 2x=2x\] that is your numerator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{3x+7}{2x}\times 2x = 3x+7\] that is why you did it to get rid of the pesky 2x in the bottom

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0of course you have to multiply the whole denominator by 2x, not just part of it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you numerator is 2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your denominator is 3x+72x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which give a denominator of x+7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so our "final answer" is \[f(g(x))=\frac{2x}{x+7}\]

moongazer
 5 years ago
Best ResponseYou've already chosen the best response.0i just got confused i forgot that 3x and 2x can be combined

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah but clear now yes?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah but clear now yes?
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