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I have been searching online for a method that shows whether or not a function has an inverse and the fastest method I can find is that you can prove that $f(x)=f(y)\Longrightarrow x=y$ which proves that the function is one-to-one. For the function $y=\frac{x}{x^2+1}$ specifically, what is the fastest way to prove that this function doesn't have an inverse.

  • Don't functions have domains and codomains? – Angina Seng Jun 22 '19 at 05:59
  • What does that mean? –  Jun 22 '19 at 06:00
  • @user532874 For a function to be invertible, it must be onto. This means that the codomain of the function, which must be defined when defining the function must be the same as its range, which depends on the domain of the function. Both domain and codomain must be defined when defining the function. For example, $\frac{x}{x^2+1}$ is invertible if its domain is ${0}$ and its codomain is ${0}$, but it is not invertible if its domain and codomain are $\mathbb{R}$. – Amit Rajaraman Jun 22 '19 at 06:03
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    Maybe add what your background knowledge is; I assume we are talking on a precalculus level here, and "function" for you is a formula sending real numbers to real numbers ... If you know calculus, well, a continuous function needs to be either increasing or decreasing on its entire domain to have an inverse. With standard curve sketching, you'll see that your function changes from decreasing to increasing to decreasing again. – Torsten Schoeneberg Jun 22 '19 at 06:04
  • @TorstenSchoeneberg I'm learning calculus, how do I prove if the function is increasing or decreasing on the entire domain? I'm guessing we use the derivative but how exactly? –  Jun 22 '19 at 06:37
  • Couldn't you have points where the derivative is zero and the function is still one-to-one? The derivative doesn't have to be completely positive or completely negative right? –  Jun 22 '19 at 06:39
  • Yes, one can have such points (e.g. $h(x)=x^3$ is one-to-one, despite $h'(0)=0$). That is why in my comment I talked about in-/decreasing behaviour, not critical numbers. Remember that every change of increasing to decreasing or vice versa happens at a critical number, but not at every critical number such a change happens necessarily (as in the above $h$). In your example, however, the sign of the derivative does change at the critical $x=\pm1$, so the function is not one-to-one. – Torsten Schoeneberg Jun 22 '19 at 17:33

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Your function $f(x)=\frac{x}{x^2+1}$ is of course continuous, since it is a ratio of polynomials and the denominator has no real roots. If a continuous function is a bijection (i.e. it has an inverse), then it must be monotonic (see this question for a proof). But $f(x)$ is clearly not monotonic, since $f(0)<f(1)$ but $f(2)<f(1)$.

YiFan Tey
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$f(\frac 1x)=\dfrac{\frac 1x}{\frac 1{x^2}+1}=\dfrac{x^2}{x(1+x^2)}=\dfrac{x}{1+x^2}=f(x)\quad$ thus $f$ is not injective on whole $\mathbb R$.

You need to restrict the function to intervals $(-\infty,-1),[-1,1],(1,+\infty)$ to recover injectivity (that need to be confirmed by studying $f'(x)$ sign on these intervals and show $f$ is strictly monotonic on these).

zwim
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