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This may seem like an elementary question, but bear with me; you'll find that it is actually quite hard. Consider the function $$f(x)=\frac{a_nx^n}{\sum_{i=0}^n a_ix^i}$$ with all $a_i\geq 0$ and $a_0=1$. Can you prove that the maximum derivative over $x\in (0,\infty)$, if it exists, is unique?

The usual elementary root counting techniques do not work since $f''(x)$ may have many sign changes. It is hard to control the number of zeros. Perhaps arguing that the numerator and denominator of $f(x)$ and $f'(x)$ are strictly monotone increasing may help.

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  • $\begingroup$ What makes you think this is true? Take $f(x)=\frac{x^2}{x^2 + ax + 1}$ for $a$ between $1$ and $2$; the derivative has two local maxima, one positive and one negative. For $a=1.3$, the negative maximum is smaller; for $a=1.5$ the negative maximum is greater. $\endgroup$ Commented Oct 22, 2014 at 7:03
  • $\begingroup$ I am taking the maximum derivative over $x\in (0,\infty)$. $\endgroup$
    – user60751
    Commented Oct 22, 2014 at 17:24
  • $\begingroup$ oops, my mistake. $\endgroup$ Commented Oct 22, 2014 at 23:55

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This appears to be false. Consider $$ f(x)=\frac{\frac{x^{10}}{100}}{1+b+\frac{x^8}{2}+\frac{x^{10}}{100}} $$ for $b\in [5,10]$.

In this range, $f'(x)$ has two local maxima in $(0,\infty)$. One occurs at an $x$-value less than $2$ and the other occurs at an $x$-value between $3$ and $5$. At $b=5$ the second local maximum has a greater value; at $b=10$ the first local maximum has a greater value. So for some $b$ in that range the two maxima have equal value.

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  • $\begingroup$ I'd be interested if there is any reason other than looking at a graph or two that this would be a reasonable thing to think, or if there's an obvious reason that dramatic counterexamples are hard to come by (I guess we think that as $f(0)=0$ and $f(x)<1$ for positive $x$ but approaches $1$ relatively quickly, there isn't much room for dramatic maxima in $f'(x)$). $\endgroup$ Commented Mar 23, 2015 at 1:03

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