# Roots of a sum of functions

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?

• 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 $g(x_2)$? Oct 9, 2020 at 0:09
• 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... Oct 9, 2020 at 0:28
• 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. Oct 9, 2020 at 2:57
• 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. Oct 9, 2020 at 3:02
• 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. Oct 9, 2020 at 3:45