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 $$y = \frac3{x+1}$$
Switch places of x and y $$x= \frac3{y+1}$$
Try to solve for y. So I multiply the denominator by x to get rid of it $$x(y+1) = 3$$
After multiplying, I'm left with $$xy + x = 3$$
Which then converts to $$2xy = 3$$
Then I get rid of
2xon the left, placing it on the right $$y = 3 - 2x$$Now I convert
yto 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!
$\endgroup$ 23 Answers
$\begingroup$After step 2. you could just divide by $x$ to get $$y+1 = \frac3x$$ and then subtract $1$ to get $$y = \frac3x - 1 = \frac{3-x}x$$
$\endgroup$ 2 $\begingroup$Step 4 is wrong, check it. instead, divide by x after step 2, and carry over 1.
$\endgroup$ 3 $\begingroup$$XY+X=3 \\XY=3-X \\Y=(3-X)/X$
This is correct steps after step $3$
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