Is it possible to get inverse of all be functions? For example, can we calculate inverse of $y=x^3+x$?

  • $\begingroup$ Do you ask if there exists an inverse or how to find it in a closed form? For this function you can do that using the Cardano-Tartaglia method - en.wikipedia.org/wiki/Cubic_equation $\endgroup$ – Dennis Gulko Nov 20 '12 at 19:00

In general: no!

You need a function to be one-to-one for there to be any hope of an inverse. Take for example the function $f(x) = x^2.$ This function is two-to-one: $f(\pm k) = k^2.$ Let's assume we're working over the real numbers. What is the inverse of $-1$? Well, there isn't one. What is the inverse of $4$? Well, it could be either $-2$ or $+2$ because $(-2)^2 = 2^2 = 4.$ (The inverse is not well defined.)

In general: if a function is one-to-one then there is a well defined inverse from the range of the function back to the domain.

In your example, you have $f(x) = x^3 + x.$ This function is one-to-one because $f'(x) \neq 0$ for all $ x \in \mathbb{R}$. The range of $f$ is the whole of the real numbers, so there will be a well defined inverse $f^{-1} : \mathbb{R} \to \mathbb{R}.$ However, it is difficult to actually write the inverse down. It will be some complicated expression. Although, we can say for certain is that it exists.


The function $f$ with $f(x)=x^3+x$ has an inverse that can be stated with an explicit formula that only uses elementary arithmetic and radicals. Here is a formula for that example's inverse function. The markup isn't working because of the carat in the url. Copy and paste the whole thing:


But that is special to the case where you have an invertible cubic polynomial. In general, when $f$ is an invertible function defined on some subset of $\mathbb{R}$, then only "formula" that you will have for its inverse is notational: $f^{-1}(x)$.

  • $\begingroup$ what the inverse of f(x) in this example? $\endgroup$ – 2 8 Nov 20 '12 at 19:05

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