Changing the subject of a formula to find the inverse function

NeedingWD40

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Jan 14, 2019
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Hi, I am stuck with a problem:

f(x) = x/(x+1) , find the inverse function. I know to let y=x/(x+1), change the subject of the formula to y, then replace x with y and y with x, but I'm stuck at the changing the subject stage....

Many thanks in advance for your time.
 
Okay, making that change, we have:

[MATH]x=\frac{y}{y+1}[/MATH]
What do we get if we multiply both sides by \(y+1\) ?
 
Okay, making that change, we have:

[MATH]x=\frac{y}{y+1}[/MATH]
What do we get if we multiply both sides by \(y+1\) ?
... I started down that route and got x(y+1) = y, but that multiplies out to xy + x = y, which seemed to get me no closer to y by itself, but...

now I see if I bring 'all the ys to one side'... x = y - xy, then factorise x = y(1 - x), then divide both sides by (1 - x)...

x/(1-x) = y ... I have the inverse function!

I think I just needed encouragement that I was on the right track and to leave it and come back and look at it again ?

Many thanks!

(and I really should learn to use the math formatting!)
 
Bravo!

We can check your work by ensuring [MATH]y\left(y^{-1}(x)\right)=y^{-1}\left(y(x)\right)=x[/MATH]:

[MATH]y\left(y^{-1}(x)\right)=\frac{\dfrac{x}{1-x}}{\dfrac{x}{1-x}+1}=\frac{x}{x+1-x}=x\quad\checkmark[/MATH]
[MATH]y^{-1}\left(y(x)\right)=\frac{\dfrac{x}{x+1}}{1-\dfrac{x}{x+1}}=\frac{x}{x+1-x}=x\quad\checkmark[/MATH]
So, it all checks out. :)
 
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