## moongazer 5 years ago 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)

1. anonymous

step on write $f(g(x))$

2. anonymous

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

3. anonymous

so far so good?

4. moongazer

yup

5. anonymous

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

6. anonymous

zat ok?

7. anonymous

is jsut an exercise in simplication

8. anonymous

simplification

9. anonymous

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

10. moongazer

yup that is what my teacher told me

11. 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

12. 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 $\frac{1}{-2}$

13. anonymous

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

14. moongazer

what??

15. anonymous

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

16. moongazer

what happened to the ( )^2?

17. anonymous

oops i lost you

18. anonymous

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

19. moongazer

yes because i think it is easier

20. anonymous

ok in that case there is no square in the denominator

21. moongazer

ooooohhh now i know

22. anonymous

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

23. anonymous

if so then everything i wrote above works yes?

24. anonymous

if not let me know

25. moongazer

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

26. moongazer

am i correct?

27. anonymous

no.

28. moongazer

why?

29. anonymous

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

30. anonymous

not $3x+7$

31. anonymous

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

32. moongazer

yup i forgot to type 2x sorry for that

33. moongazer

34. anonymous

ok good

35. anonymous

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

36. anonymous

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

37. anonymous

k?

38. moongazer

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

39. anonymous

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

40. anonymous

$\frac{3x+7}{2x}\times 2x=3x+7$

41. moongazer

ohhh ok

42. 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

43. anonymous

clear or no?

44. moongazer

wait ill do it on a paper ^_^

45. anonymous

k

46. moongazer

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

47. anonymous

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

48. moongazer

ok

49. anonymous

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

50. anonymous

$1\times 2x=2x$ that is your numerator

51. moongazer

ok

52. 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

53. anonymous

$-1\times 2x=-2x$

54. moongazer

ok

55. anonymous

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

56. anonymous

so you numerator is 2x

57. anonymous

58. anonymous

which give a denominator of x+7

59. moongazer

ok

60. anonymous

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

61. moongazer

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

62. anonymous

ah but clear now yes?

63. moongazer

yes! thanks^_^

64. anonymous

welcome

65. anonymous

ah but clear now yes?