# Inverse of function, containing a fraction

This is basic, I know, but I cannot seem to come up with the right answer.

Find the inverse of the function: $$f(x)= \frac3{x+1}$$

My steps: 1. Convert f(x) to y $$y = \frac3{x+1}$$

1. Switch places of x and y $$x= \frac3{y+1}$$

2. Try to solve for y. So I multiply the denominator by x to get rid of it $$x(y+1) = 3$$

3. After multiplying, I'm left with $$xy + x = 3$$

4. Which then converts to $$2xy = 3$$

5. Then I get rid of 2x on the left, placing it on the right $$y = 3 - 2x$$

6. Now I convert y to the inverse function $$f^{-1}(x) = 3 - 2x$$

My answer is obviously wrong. The correct answer is: $$f^{-1}(x) = \frac{3-x}{x}$$

Where did I mess up?

Thanks!

• Step 4: $xy+x\neq 2xy$ Feb 4, 2014 at 20:07
• Man oh man... what a day. Thanks! :) Feb 4, 2014 at 22:43

After step 2. you could just divide by $x$ to get $$y+1 = \frac3x$$ and then subtract $1$ to get $$y = \frac3x - 1 = \frac{3-x}x$$

• Thanks for this! I see where I messed up (xy + x should have then turned into y = 3-x/x). What is the process to "divide by x"? It sounds like I might be doing extra unnecessary steps. Feb 4, 2014 at 22:42
• @WesFoster It's multiplication of both sides by $\frac1x$, of course assuming $x\ne 0$, but since $f(y) \ne 0 \quad\forall\ y\in\mathbb R$, this is not a problem. Feb 5, 2014 at 21:43

Step 4 is wrong, check it. instead, divide by x after step 2, and carry over 1.

• This is not a answer. Is a comment. Feb 4, 2014 at 20:11
• The question was "where did I mess up?" I answered that. Feb 4, 2014 at 20:14
$XY+X=3 \\XY=3-X \\Y=(3-X)/X$
This is correct steps after step $3$