Inverse function of $x-\lfloor x \rfloor $ and $(x-\lfloor x \rfloor)^2$

Question

I need to find the inverse of these two functions if they exist:

$$f_1 = x-⌊x⌋, 1\leq x<2, 0\leq f_1<1$$ and $$f_2 = (x-⌊x⌋)^2, 1\leq x<2, 0\leq f_2<1$$

I worked through it and I think they are both bijective functions, but I am not exactly sure, and Having trouble finding the actual inverse function for both these two. Do these two have inverse functions and how do you find them?

Answer 1

Hint: if $1\leq x < 2$, then $\lfloor x\rfloor$ is always the same number for all those $x$'s. If you write this number instead of $\lfloor x \rfloor$ in your formulas, you'll get two easy-to-invert functions.

Answer 2

$f_1(x)=x-\lfloor x\rfloor$ is called the fractional part function. It returns the value after the decimal point in $x$, and is denoted as $\{x\}$. Note that while $f_1(x)$ is many-one over $\Bbb R~(f_1(x+1)=f_1(x))$, its restriction to $[1,2)$ is injective and onto over $[0,1)$.

In the given domain, $\lfloor x\rfloor=1$, so that $f_1(x)=x-1$. The inverse of this is easily calculated as $f_1^{-1}(x)=y+1,0\le y<1$. Similarly, the inverse of $f_2(x)=\{x\}^2=(x-1)^2$ is $f_2^{-1}(x)=1+\sqrt y,0\le y<1$.