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

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?

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.
$$f_1(x)=x-\lfloor x\rfloor$$ 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 $$f_2^{-1}(x)=1+\sqrt y,0\le y<1$$.