What is the pre-image of $f(x) = \frac{1 }{1 + x^2}$ at $\{0\}$

Question

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?

Answer 1

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

Answer 2

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: $f(D) = (0,1],$

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