2
$\begingroup$

R to R

$f(x) = \lfloor \frac{x-2}{2} \rfloor $

If $T = \{2\}$, find $f^{-1}(T)$

Is $f^{-1}(T)$ the inverse or the "image", and how do you know that we're talking about the image and not the inverse?

There shouldn't be any inverse since the function is not one-to-one, nor is it onto since it's $\mathbb{R}\to\mathbb{R}$ and not $\mathbb{R}\to\mathbb{Z}$.

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    What is $T$? Also, $f^{-1}(x)$ is never a notation for the image. Typically, you would denote the image by Im($f$). A more likely notational problem with $f^{-1}$ would be confusion with the reciprocal $1/f(x)$.2012-12-07
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    T is the domain of the variable2012-12-07
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    Ah, your new edit clears up your notational confusion. $f^{-1}(x)$ is notation for the inverse function applied to the variable $x$. $f^{-1}(T)$ is notation for the *pre*image of the set $T$ under $f$, i.e. $\{x\in \mathbb{R}\,|\,f(x)\in T\}$. Notice that the difference is whether the argument to $f^{-1}$ is an element of the codomain (or a variable standing for such) or a subset of the codomain.2012-12-07

3 Answers 3