Question

what is the inverse of this function
i(x)=3-4/2x+1
Please help
Can anybody help me please I am stressed to the max LOL

Answer 1

\[i(x)=3 -\frac{4}{2y+1}\]is the same as saying
\[x=3-\frac{4}{2y+1}\]
What you're trying to find in an inverse is basically switching the positions of the x and y values and making the x-value the subject again.
So...
\[y=3-\frac{4}{2x+1}\]
Now we make x the subject. Try and do that and someone else may get back to you as I have to leave. Sorry for the inconvenience.

Answer 2

the last equation you have, is that what you want the inverse of because that is the same as what you have in the original question.

Answer 3

i want the inverse of i(x)=\[3-4\div2x+1\]

Answer 4

ok, so in that equation, switch the x and y. so you get \[x=3-\frac{ 4 }{2y+1}\]

Answer 5

then solve for y. so what is the first thing you will do?

Answer 6

x-3

Answer 7

correct. so then you have x-3=4/(2y+1). Now what you want to do is get y out of the bottom

Answer 8

isn't the rhs negative? Do I then divide by y?

Answer 9

yes it is, sorry i thought i put the sign in there...you want to multiply both sides by (2y+1) so that you get (2y+1)(x-3)=4

Answer 10

negative 4, sorry again

Answer 11

now that you have (2y+1)(x-3)=-4 how do you get to y=?

Answer 12

do i expand the brackets?

Answer 13

no, the first step is to get (x-3) to the rhs

Answer 14

so divide by x-3

Answer 15

yes, and then all you have left is 2y+1=-4/(x-3)

Answer 16

so then -1 and get 2y = (-4/(x-3)) - 1 ?

Answer 17

yes, and then divide everything by 2

Answer 18

\[y= -2/((x-3)/2)- 1/2\]

Answer 19

I think that is right, just to be sure, I got \[y=\frac{ -2 }{ x-3 }+1/2\]

Answer 20

is this for written out hw or online hw?

Answer 21

so the denominator is not divided by 2 as well?

Answer 22

written out :)

Answer 23

no because it is a fraction divided by a number, it is really \[(\frac{ 1 }{ 2 })(\frac{ -4 }{ x-3 })\]

Answer 24

ok, and have you ever heard of wolframalpha.com?

Answer 25

yes but I needed step by step to help me understand the process

Answer 26

ok, you might want to check it against what my answer was just to be sure, and if there is a different answer, let me know and ill figure it out

Answer 27

No I am happy with this as my asymptotes now match :) Thank you so very very much xoxo

Answer 28

ok, no problem. Hope you do well on your hw!!

Answer 29

Thanks again :)