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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?

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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.

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  • $\begingroup$ Thank you so much! $\endgroup$ – Pengibaby Feb 5 at 3:35
  • $\begingroup$ You're very welcome :) $\endgroup$ – eudes Feb 5 at 3:48
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$f_1(x)=x-\lfloor x\rfloor$ 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 $f_2^{-1}(x)=1+\sqrt y,0\le y<1$.

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  • $\begingroup$ Thank you for the explanation and answer :) $\endgroup$ – Pengibaby Feb 5 at 4:10

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