1. ## Inverse Functions

I'm having the hardest time finding the inverse function of

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

I know I exchange all the the x's for y's and solve for y but I'm having trouble doing so.... Help

2. Try long dividing first.

This gives $\frac{x+1}{2x+1}=\frac{1}{2(2x+1)}+\frac{1}{2}$

$x=\frac{1}{2(2y+1)}+\frac{1}{2}$

$x-\frac{1}{2}=\frac{1}{2(2y+1)}$

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

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

$y=\frac{1-x}{2x-1}$

3. Originally Posted by galactus
Try long dividing first.

This gives $\frac{x+1}{2x+1}=\frac{1}{2(2x+1)}+\frac{1}{2}$

$x=\frac{1}{2(2y+1)}+\frac{1}{2}$

$x-\frac{1}{2}=\frac{1}{2(2y+1)}$

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

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

$y=\frac{1-x}{2x-1}$
i never thought of attempting a problem like this that way.

Here's the method i usually use.

$y = \frac {x + 1}{2x + 1}$

For inverse, switch x and y and solve for y

$\Rightarrow x = \frac {y + 1}{2y + 1}$

$\Rightarrow 2xy + x = y + 1$

$\Rightarrow 2xy - y = 1 - x$

$\Rightarrow y(2x - 1) = 1 - x$

$\Rightarrow y = \frac {1 - x}{2x - 1}$

4. Your method appears a little more efficient. Just another way to go about it.

5. Thanks alot guys!!