moongazer
  • moongazer
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
step on write \[f(g(x))\]
anonymous
  • anonymous
step two replace \[g(x)\] by \[\frac{3x+7}{2x}\] to get \[f(\frac{3x+7}{2x})\]
anonymous
  • anonymous
so far so good?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

moongazer
  • moongazer
yup
anonymous
  • 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
  • anonymous
zat ok?
anonymous
  • anonymous
is jsut an exercise in simplication
anonymous
  • anonymous
simplification
anonymous
  • 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
  • moongazer
yup that is what my teacher told me
anonymous
  • 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
  • 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
  • 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
  • moongazer
what??
anonymous
  • anonymous
then multiply top and bottom by 2x to get \[f(g(x))=\frac{2x}{3x+7-2x}=\frac{2x}{x+7}\]\][
moongazer
  • moongazer
what happened to the ( )^2?
anonymous
  • anonymous
oops i lost you
anonymous
  • anonymous
are we going to replace \[f(x)=\frac{x+1}{x^2-1}\] by \[f(x)=\frac{1}{x-1}\]?
moongazer
  • moongazer
yes because i think it is easier
anonymous
  • anonymous
ok in that case there is no square in the denominator
moongazer
  • moongazer
ooooohhh now i know
anonymous
  • anonymous
if we make that change in f, then \[f(g(x))=f(\frac{3x+7}{2x})\] yes?
anonymous
  • anonymous
if so then everything i wrote above works yes?
anonymous
  • anonymous
if not let me know
moongazer
  • moongazer
isn't it it should first be 1/(3x+7)- 1?
moongazer
  • moongazer
am i correct?
anonymous
  • anonymous
no.
moongazer
  • moongazer
why?
anonymous
  • anonymous
\[g(x)=\frac{3x+7}{2x}\]
anonymous
  • anonymous
not \[3x+7\]
anonymous
  • anonymous
so when we replace x in f(x) by \[\frac{3x+7}{2x}\] we get \[\frac{1}{\frac{3x+7}{2x}-1}\]
moongazer
  • moongazer
yup i forgot to type 2x sorry for that
moongazer
  • moongazer
please continue
anonymous
  • anonymous
ok good
anonymous
  • anonymous
now you can simplify this compound fraction by multiplying top and bottom by 2x
anonymous
  • 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
  • anonymous
k?
moongazer
  • moongazer
what happened with the 2x below 3x+7?
anonymous
  • anonymous
i multiplied top and bottom by 2x to get rid of it. canceled from the denominator
anonymous
  • anonymous
\[\frac{3x+7}{2x}\times 2x=3x+7\]
moongazer
  • moongazer
ohhh ok
anonymous
  • 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
  • anonymous
clear or no?
moongazer
  • moongazer
wait ill do it on a paper ^_^
anonymous
  • anonymous
k
moongazer
  • moongazer
how did it become 2x/2x+7?
anonymous
  • anonymous
ok let us start with \[\frac{1}{\frac{3x+7}{2x}-1}\] ok?
moongazer
  • moongazer
ok
anonymous
  • anonymous
multiply top and bottom by 2x. what do you get?
anonymous
  • anonymous
\[1\times 2x=2x\] that is your numerator
moongazer
  • moongazer
ok
anonymous
  • 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
  • anonymous
\[-1\times 2x=-2x\]
moongazer
  • moongazer
ok
anonymous
  • anonymous
of course you have to multiply the whole denominator by 2x, not just part of it.
anonymous
  • anonymous
so you numerator is 2x
anonymous
  • anonymous
your denominator is 3x+7-2x
anonymous
  • anonymous
which give a denominator of x+7
moongazer
  • moongazer
ok
anonymous
  • anonymous
so our "final answer" is \[f(g(x))=\frac{2x}{x+7}\]
moongazer
  • moongazer
i just got confused i forgot that 3x and -2x can be combined
anonymous
  • anonymous
ah but clear now yes?
moongazer
  • moongazer
yes! thanks^_^
anonymous
  • anonymous
welcome
anonymous
  • anonymous
ah but clear now yes?

Looking for something else?

Not the answer you are looking for? Search for more explanations.