Suppose I have a vector function $f(x): \mathbb{R}^n \rightarrow \mathbb{R}^n$, defined as the elementwise application of some real function. I know that this real function is strictly increasing and has image $\mathbb{R}$. The function f(x) thus has only a single root.

How can I show that adding a linear function to $f(x)$ resulting in $g(x) = f(x) + Mx$ will also give a function with a single root on the domain? Also assume that $f(x) \neq -Mx$.

In general, does adding a linear function to some function can increase the number of roots above one?

  • $\begingroup$ I am unfamiliar with the phrase "only one root in the domain". Therefore, I advise that you take the following idea with a large grain of salt, because the idea may well lead nowhere. If $x_1 \neq x_2$ with $f(x_1) \neq f(x_2)$, is it possible for $g(x_1)$ to equal $g(x_2)$? $\endgroup$ Oct 9, 2020 at 0:09
  • $\begingroup$ Seems to be a very good idea. An example of function would be $f(x) = x+ \sin(x)$, for which $g(x) =\sin(x)$ can have multiple roots. I might have to use Taylor series for this... $\endgroup$
    – Undead
    Oct 9, 2020 at 0:28
  • $\begingroup$ What does "strictly increasing on some domain" mean? The domain of $f$ is $\mathbb R^n$, so it is either strictly increasing on its domain, or just on some subset of the domain. If it's the latter, you can't conclude that it has only one root. $\endgroup$ Oct 9, 2020 at 2:57
  • $\begingroup$ To answer your question, adding a linear function to some function with 1 root will not in general increase the number of roots. Take any polynomial $f(x)=ax+b$ with $a\neq 0$. There are infinitely many linear functions $h(x)=cx+d$ such that $f(x)+h(x)$ still has 1 root. $\endgroup$ Oct 9, 2020 at 3:02
  • $\begingroup$ OK I rephrased the question, forget about "some" domain (by that I meant subset of the domain) and let's say it's the domain. $\endgroup$
    – Undead
    Oct 9, 2020 at 3:45


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