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Suppose a function

$$ g(x) = a\cdot q(x) + b\cdot f(x) $$ where $q(x)$ is an arbitrary quadratic function, $f(x)$ is an arbitrary function that's decreasing and $f(x) \leq 0$ for all $x$.

Is it possible to show that $g(x)$ only has 0,1,or 2 roots? Here a and b are arbitrary positive scalars.

Semiclassical
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NavStoke
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  • If $a,b$ are positive, then you can assume $a+b=1$ without loss of generality. (Let $a'=a/(a+b)$, $b'=b/(a+b)$, then note that $g(x)/(a+b)$ has the same roots as $g(x)$ itself.) – Semiclassical Mar 18 '24 at 20:49
  • What have you tried? What do you think? Could we make $ g(x) =0 $, say on $ x \geq 0$? – Calvin Lin Mar 18 '24 at 21:17
  • If $b > 0$, then $a,b$ can be dropped from the question, as $aq$ and $bf$ are also functions meeting the exact same conditions $q$ and $f$ itself. If $b$ is allowed to be $< 0$, then $a,b$ can also be dropped, but with the change that $f$ is only required to be monotone, and can be either $\le 0$ everywhere or $\ge 0$ everywhere. – Paul Sinclair Mar 19 '24 at 19:03
  • Have you played around with specific functions for $f$ and $q$ with the idea of seeing if you can find a way to have more than $2$ roots? Say let $q(x) = x^2$, and maybe just set $f(x) = 0$ for $x < 0$, then figure out what you could do for $x > 0$ that would result in more roots in $g$? – Paul Sinclair Mar 19 '24 at 19:07

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