Following my question on Meta, I post this as a new question; it was originally asked by Gmgfg, and closed due to the lack of context, background, and shown effort.

Let $f\colon[a,b]\to\mathbb{R}$ be a function continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)=0$. Show that there exists $c\in(a,b)$ such that $f'(c)+f(c)=f(c)f'(c)$.

Now, given the assumptions this looks like it should be a straightforward application of Rolle's theorem. However:

  • as far as I can tell there is no simple auxiliary function $\Phi$ such that $\Phi'= f+f'-f\cdot f'$ to which one could apply Rolle's theorem. (If there is one, I failed to find it.)

  • indeed, while the RHS could come from $\left(\frac{f^2}{2}\right)'$; it's mostly the LHS which looks difficult to handle (and there is no clear advantage I can see in introducing an antiderivative $F$ of $f$ to have it be $(F+f)'$).

I verified the statement for some functions I could think of, such as $x\mapsto x(1-x)$ and $x\mapsto \sin \pi x$ on $[a,b]=[0,1]$. So, in that regard, it seems to hold at least against basic sanity checks.

However, what bothers me is the lack of "homogeneity." Usually, in things like that I'm used to saying that "without loss of generality, one can assume $[a,b]=[0,1]$." It does not appear to be the case here: if, given $f$, one defines $g\colon[0,1]\to\mathbb{R}$ by $g(x) = f((b-a)x+a)$, finding $c\in(0,1)$ such that $g'(c)+g(c)=g(c)g'(c)$ does not directly yield $c'$ such that $f'(c')+f(c')=f(c')f'(c')$ (but rather would give $c'$ such that $f'(c')+(b-a)f(c')=f(c')f'(c')$, if I'm not mistaken).

So, in short: is it true? And how to prove or disprove it — I'm at a loss, and embarrassed about it.


1 Answer 1


Let $$ g(x) = f(x)e^{x-f(x)}. $$ Then $g(a) = g(b) = 0$ since $f(a)=f(b)=0$, so $g'(c) = 0$ for some $c\in(a,b)$. However, $$ g'(x) = f'(x)e^{x-f(x)} + f(x)e^{x-f(x)}(1-f'(x)) = e^{x-f(x)}(f'(x)+f(x)-f(x)f'(x)),$$ so $g'(c) = 0\implies f'(c)+f(c)-f(c)f'(c) = 0$, i.e. $f'(c)+f(c) = f(c)f'(c)$.

  • 3
    $\begingroup$ Wow, this was fast -- also, this auxiliary function does look really contrived to me. Any intuition as to how you came up with it? $\endgroup$
    – Clement C.
    Mar 31, 2017 at 1:47
  • 1
    $\begingroup$ I guess what you could do is consider something of the form $f(x) e^{g(x)} $ and see what function $g$ does the trick. Following this idea gives $g'(x) = 1-f'(x)$ in this case. $\endgroup$ Mar 31, 2017 at 1:50
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    $\begingroup$ @ClementC. essentially I was trying to solve $f'+f = ff'$ via integrating factors to get the right function that derived correctly. So I first multiplied by $e^x$ on both sides to get $y'$ on the LHS, where $y(x) = e^xf(x)$. But then I noticed that the RHS was just $y(x)f'(x)$, and the integrating factor for $y'-f'y$ is $e^{-f(x)}$. That suggested I should consider $e^{-f(x)}y(x) = e^{-f(x)}e^xf(x) = f(x)e^{x-f(x)}$. $\endgroup$
    – Joey Zou
    Mar 31, 2017 at 1:53
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    $\begingroup$ @JoeyZou Great. Thank you! $\endgroup$
    – Clement C.
    Mar 31, 2017 at 1:59
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    $\begingroup$ your comment is more important than the answer. +1 for both the comment and the answer. $\endgroup$
    – Paramanand Singh
    Apr 1, 2017 at 4:07

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