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I consider the function $f(x) = x^2 : \mathbb{R} \rightarrow \mathbb{R}$ whose image is $[0, + \infty)$. For the sake of simplicity: domain $D = \mathbb{R}$, codomain $C = \mathbb{R}$.

If I consider $A = [-25, 25]$ subset of the codomain $C$, this subset contains elements $[-25, 0)$ which don't have a corresponding element in the domain $D$. In this case is it possible to evaluate the inverse image of $A$? I tried to do it in this way.

According to the definition of the inverse image:

$$ f^{-1}(A) = \lbrace x \in D : f(x) \in A \rbrace $$

$$ f^{-1}(A) = f^{-1}([0, 25]) = \lbrace x \in [-5, 5] \rbrace $$

Is it correct?

EDIT: the Mathematica software gave me the same result I wrote.

Gennaro Arguzzi
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1 Answers1

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$A=[-25,25]$ is not a subset of the $f$'s range $[0,\infty),$ so $A$ does not have a preimage under $f.$

$f$'s range (let's call it $R$) is a subset of $f$'s codomain, and $R$'s preimage is $f$'s domain.

ryang
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