Question

inverse f(x)= (2x+1) / (3x+2)

Answer 1

\[f(x)=(2x+1)\div (3x+2) \] right?

Answer 2

yeah
the inverse

Answer 3

of that please

Answer 4

If so it would be (3x+2)(2)-( 2x-1)(3) / \[(3x+2)^{2}\]

Answer 5

do you mean the inverse function for
\[ \Large f(x)=\frac{2x+1}{3x+2} \]

Answer 6

yes ^

Answer 7

yes
but thats not the anserw but the first step I did

Answer 8

ok then you have to solve for x the equation
\[ \Large y=f(x)=\frac{2x+1}{3x+2} \]
tell me what you get

Answer 9

would the anserw be 6 over \[(3x+2)^{2}\]

Answer 10

No @hooverst

Answer 11

@sabriiina. are you working on this?

Answer 12

how would u do it? I probably missed a step

Answer 13

i know what the answer is

Answer 14

i don't know how to get it

Answer 15

im stuck at the step
y= 2y+1 -2x
-------------
3x

Answer 16

ok let me try
\[ \Large y=\frac{2x+1}{3x+2} \]
\[ \Large y(3x+2)=2x+1 \]
\[ \Large 3xy+2y=2x+1 \]
\[ \Large 3xy-2x=1-2y \]
\[ \Large x(3y-2)=1-2y \]
\[ \Large x=\frac{1-2y}{3y-2} \]
then
\[ \Large f^{-1}(y)=\frac{1-2y}{3y-2} \]

Answer 17

do you get this?

Answer 18

\[y=\frac{2x+1}{3x+2}\]
\[({3x+2})y={2x+1}\]
\[{3xy+2y}={2x+1}\]
\[3xy-2x=1-2y\]
\[x(3y-2)=1-2y\]
\[x=\frac{1-2y}{3y-2}\]

Answer 19

thank you. i get it now :)