Inverse of function, containing a fraction

Question

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!

Answer 1

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

Answer 2

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