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?
Best Answer
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: $f(D) = (0,1],$
$0\not\in f(D).$