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I need to find the pre-image at $\{0\}$ with the funtion in the domain $$f: \mathbb{R} \to \mathbb{R}$$

$$f(x) = \frac{1 }{1 + x^2}$$

I first found the inverse function $$f^{-1} = y = \frac{1 }{1 + x^2}$$ $$x = \frac{1 }{1 + y^2}$$ $$1 + y^2 = \frac{1 }{x}$$ $$y = (\frac{1 }{x} - 1)^{\frac{1 }{2}} = f^{-1}(x)$$

Then I replace $x$ with $0$: $$ f^{-1}({0}) = (\frac{1 }{0} - 1)^{\frac{1}{2}} $$

But since you can't divide by $0$, their is no pre-image at $0$ for this function in the given domain and range. Am I correct?

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  • $\begingroup$ $f$ is not injective (for example $f(-1) = f(1)$) so $f$ does not have an inverse function. $\endgroup$ – Daniel Schepler Nov 1 '17 at 19:51
  • $\begingroup$ So that means that their is no pre-image at 0? $\endgroup$ – user496674 Nov 1 '17 at 19:53
  • $\begingroup$ yes it does but its not a function $\endgroup$ – J. Sadek Nov 1 '17 at 19:55
  • $\begingroup$ Then how would one find that? Because I've only learned how to find the pre-image when their is an inverse function. $\endgroup$ – user496674 Nov 1 '17 at 19:57
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    $\begingroup$ By definition, $f^{-1}(\{0\}) = \{ x \in \mathbb{R} : f(x) \in \{ 0 \} \}$. $\endgroup$ – Daniel Schepler Nov 1 '17 at 20:01
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A bit of support to Sadek, rephrased:

$\dfrac{1}{1+x^2} = 0$

has no solution for $x \in \mathbb{R}.$

$f(x) = \dfrac{1}{1+x^2} : $

Domain: $D= \mathbb{R}.$ Range: $f(D) = (0,1],$

$0\not\in f(D).$

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$\frac 1 {1+x^2}=0$ is imposible (because it gives $1=0$) therefore $f^{-1}(\{0\})=\emptyset$

again $f^{-1}$ is not the inverse function of $f$ it is the inverse image of $f$ and takes a set as an argument and gives back a set

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    $\begingroup$ "$0$ does not have a pre-image" Yes, it does. You wrote it yourself. It's $\emptyset$. $\endgroup$ – Ennar Nov 1 '17 at 21:35
  • $\begingroup$ I'm sorry you're absolutly right. English is not my first mathematical language. In french elements of the pre-image of {y} are called the "antécédents" of y. Do they have a name in english? $\endgroup$ – J. Sadek Nov 1 '17 at 22:06
  • $\begingroup$ Not that I'm aware of. $\endgroup$ – Ennar Nov 2 '17 at 5:38

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