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This is related directly to another question about this: Continuous Functions

My question though revolves around how to approach the question in a "thinking" way. So mechanically looking at the expression one would take the square root of both sides and end up with:

$$f(x) = \pm x$$

But this doesn't really say anything about how to interpret the result. So in the other post the solutions revolved around looking at the behaviour of $x$ between $(-\infty,0)$ and $(0,\infty)$ and applying the intermediate value theorem. But what is accomplished by showing that $f(x)$ does not change sign? and also how do you arrive at the second set of solutions? Specifically I'm aware of the relationship $\sqrt{x^{2}} = |x|$, but knowing the relationship doesn't mean I know how to apply it correctly.

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    $\begingroup$ We know, for instance, that $f(1)=\pm1 $. Say it is $+1$. Then we know that $f(x)=x$ for all $x≥0$ because the function can not change sign in that region. Similarly, if $f(1)=-1$ we know that $f(x)=-x$ for all $x≥0$. $\endgroup$ – lulu Jun 23 '20 at 18:04
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There are $4$ such functions. Namely $x\mapsto\pm x$ and $x\mapsto\pm|x|$. This follows from the method suggested by @lulu in the comments. We know that $f(0)=0$ from the defining equation. Then consider some $\epsilon\gt0$. We know that $(f(\epsilon))^2=\epsilon^2$ and hence either $f(\epsilon)=\epsilon$ or $f(\epsilon)=-\epsilon$. Using the continuity of $f$ and applying the same reasoning for any other $x\gt0$ gives $f(x)=x$ or $f(x)=-x$ for all $x\gt0$ in each case respectively. Similarly we have either $f(x)=x$ or $f(x)=-x$ for all $x\lt0$. Combining these $4$ cases gives the $4$ functions originally stated.

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  • $\begingroup$ From how you explained it I gather the first ting you asked yourself is "where could the function change sign?", from there you reasoned the rest out. I could see how $\pm |x|$ would come about, but why would $\pm x$ be separate cases? Aren't they contained within the $|x|$ cases? $\endgroup$ – dc3rd Jun 23 '20 at 18:22
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    $\begingroup$ Ok I think I figured it out. I realized I put up an artificial restriction which really was holding me back. For some reason I didn't want to accept (or realize) that I can define functions piece wise.....that made it more difficult for me to reconcile everything. $\endgroup$ – dc3rd Jun 23 '20 at 18:47
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This answer might be a bit pedantic and not in the spirit of the question:

It depends on the domain of the function $f$. Let us say that the domain of $f$ is a subset of $\mathbb{C}$ called $D$, i.e., $f$ is a continuous function $D \to \mathbb{C}$ with the property that $f(x) \in \{x, -x\}$ for all $x$. We can now define the function $g$ on $D \setminus \{0\}$ by setting $g(x) = f(x)/x$. Note that $g$ is continuous and $g(x) \in \{1,-1\}$ for all $x \in D \setminus \{0\}$. Because $g$ is continuous and $\{1,-1\}$ is discrete, $g$ must be constant on every connected component of $D \setminus \{0\}$. This gives two possibilities per connected component of $D \setminus \{0\}$ (check that any combination yields a valid and distinct function $f$).


Examples:

  • If $D = \mathbb{R}$, then $D \setminus \{0\} = (-\infty,0) \cup (0,\infty)$ and you indeed get $2 \cdot 2 = 4$ possibilities.
  • If $D = \mathbb{C}$, then $D \setminus \{0\} = \mathbb{C} \setminus \{0\}$, which is connected, so in that case there are only $2$ possibilities.
  • If $D = (-1,1) \cup (3,5) \cup (7,10) \cup (12,100)$, you can check that you get $32$ possible functions $f$.
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The difference between $x$ and $-x$ is $2x$. For any nonzero $x$, this is nonzero. Hence a change of sign is not possible, except at the origin.

There are four possible combinations $(\pm\to\pm$).

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