# Is the reciprocal of f (x) = (2x+1) / (x-3) -> F^-1 (x) = (3x+1) / (x-2) ?

• Oct 29th 2013, 02:56 PM
letranger9000
Is the reciprocal of f (x) = (2x+1) / (x-3) -> F^-1 (x) = (3x+1) / (x-2) ?
I don't have access to a teacher in the present moment, I just want to make sure I understand right. If I am wrong please point out where is my error, thank you.

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

y=(2x+1) / (x-3)

x=(2y+1) / (y-3)

x(y-3)=(2y+1)

xy-3x-2y=1

xy-2y=(3x+1)

y(x-2)=(3x+1)

y(x-2)/(x-2)=(3x+1)/(x-2)

y=(3x+1)/(x-2)
• Oct 29th 2013, 03:53 PM
Soroban
Re: Is the reciprocal of f (x) = (2x+1) / (x-3) -> F^-1 (x) = (3x+1) / (x-2) ?
Hello, letranger9000!

You don't mean "reciprocal" . . . You mean "inverse".

Quote:

$\displaystyle f(x) \;=\;\frac{2x+1}{x-3}$

$\displaystyle y\;=\;\frac{2x+1}{x-3}$

$\displaystyle x\;=\;\frac{2y+1}{y-3}$

$\displaystyle x(y-3) \;=\;(2y+1)$

$\displaystyle xy-3x \;=\;2y+1$

$\displaystyle xy-2y \;=\;3x+1$

$\displaystyle y(x-2)\;=\;3x+1$

$\displaystyle y\;=\;\frac{3x+1}{x-2}$

Good work!
Punchline: /$\displaystyle f^{\text{-}1}(x) \;=\;\frac{3x+1}{x-2}$

Evaluate $\displaystyle f\left(f^{-1}(x)\right)$ or $\displaystyle f^{-1}\big(f(x)\big)$.
Either must simplify to $\displaystyle x.$

$\displaystyle f\left(f^{-1}(x)\right) \;=\;\frac{2\left(\frac{3x+1}{x-2}\right) + 1}{\left(\frac{3x+1}{x-2}\right) - 3}$

Multiply by $\displaystyle \tfrac{x-2}{x-2}\!:\;\frac{2(3x+1) + (x-2)}{(3x+1) - 3(x-2)} \;=\;\frac{6x+2 + x - 2}{3x+1-3x+6} \;=\;\frac{7x }{7} \;=\;x$

ta-DAA!
• Oct 29th 2013, 03:58 PM
HallsofIvy
Re: Is the reciprocal of f (x) = (2x+1) / (x-3) -> F^-1 (x) = (3x+1) / (x-2) ?
Be careful of your terminolgy. The reciprocal of $\displaystyle \dfrac{3x+ 1}{x- 3}$ is $\displaystyle \dfrac{x- 3}{3x+ 1}$.

But the inverse function is $\displaystyle f^{-1}(x)= \dfrac{3x+1}{x- 2}$ as you have.
• Oct 29th 2013, 04:32 PM
letranger9000
Re: Is the reciprocal of f (x) = (2x+1) / (x-3) -> F^-1 (x) = (3x+1) / (x-2) ?
Thanks guys, appreciated. And yes I meant inverse, although in french an "inverse" is a "reciprocal", hence the source of my confusion.