Calculating an inverse of a split (piecewise defined) function

Question

I am having difficulty taking the inverse of the following function (case-defined):

$$ f(x)=\begin{cases} \frac{1}{4 \sqrt{ |1-x|}} & \text{if} \ x\in [0,2] \\ 0 & \text{otherwise}\end{cases}$$

Could someone kindly explain to me how to handle the if/else portion when calculating an inverse?

I have managed to compute that if $y = f(x)$ then
$$x = 1 - \frac{1}{16y^2}$$ or
$$x = 1 + \frac{1}{16y^2}$$

However, I believe it's the last step of putting it altogether that I can't pull off.

Answer 1

The function is not $1 \text{ to } 1$; its inverse as a function does not exist. We would call it a relation instead.

Note

In the way that you have $f(x)$ defined, you need to specify $x\ne 1$. Also that function if then defined in that way on $[0,2]$ with $x\ne1$, doesn't have an inverse function . Since it's not $1 \text{ to } 1$. (Flunks the horizontal line test). The $0$ piece is trivially not $1$ to $1$ either.

Second Note

The formula you give for $x$ are correct. If you had the same function, $f(x)$, and you created two different functions,

$$ g(x) = f(x) $$ with domain $x\in[0,1)$
and $$ h(x) = f(x) $$ with domain $x\in(1,2]$

Then your formula would precisely be respective inverse functions for $g(x)$, and $h(x)$.

Answer 2

Function you provided us is not invertible, since one value of $y$ can be obtained by different values of $x$. But you can split your function into several so each part is invertible. Sometimes it's much more convenient to represent inversion graphically, since all you have to do is reflect your plot about $y = x$ line. Anyway, whatever you get is not a function, or at least classic function. It's formally called multi-valued function, but actually it's a relation.