Suppose $f\colon [a,b] \to \mathbb{R}$ is continuous and has a finite derivative $f'$ everywhere on $(a,b)$.

If $f(a)=f(b)=0$ prove that for every $y\in\mathbb{R}$ there is some $c$ in $(a,b)$ with $f'(c) = yf(c)$.

We were given a hint to apply Rolle's theorem to $h(x)f(x)$ for an $h$ that depends on $y$ but I don't understand how to apply Rolle's theorem here.

  • 1
    $\begingroup$ Please consider using LaTeX mark-up and punctutation in your posts. $\endgroup$ – Arturo Magidin Nov 23 '10 at 20:44
  • $\begingroup$ Is h(x) the same as g? $\endgroup$ – Ross Millikan Nov 23 '10 at 20:46
  • $\begingroup$ @Ross Millikan, @JimJones: almost certainly, but it was like this in the original; I only LaTeX-ed it. I'll fix that. $\endgroup$ – Arturo Magidin Nov 23 '10 at 20:50


Given $y \in \mathbb{R}$, consider the function $g_y(x)$ as below.

$g_y(x) = e^{-yx} f(x)$.

Apply Rolle's to $g_y(x)$.

(In general, if you find equations where the derivative is proportional to the function as in this case where $f'(c) \propto f(c)$, the detective in you must suspect immediately that the exponential function must be there somewhere in the picture)


Suppose you have a function $h(x)$ which is differentiable, and you consider the function $\mathcal{F}(x) = h(x)f(x)$.

Since $h$ and $f$ are both continuous on $[a,b]$, so is $\mathcal{F}$.

Since $h$ and $f$ are both differentiable on $(a,b)$, so is $\mathcal{F}$.

Since $f(a)=f(b)=0$, then $\mathcal{F}(a)=h(a)f(a)=0$ and $\mathcal{F}(b)=h(b)f(b)=0$, regardless of what $h(x)$ is.

So, $\mathcal{F}$ will also satisfy Rolle's Theorem, and so you know that there exists $c\in (a,b)$ such that $\mathcal{F}'(c)=0$.

Now, what is $\mathcal{F}'(c)$? Well, $$\mathcal{F}'(x) = h'(x)f(x) + h(x)f'(x).$$ So $\mathcal{F}'(c) = h'(c)f(c) + h(c)f'(c)$. You want $f'(c)=yf(c)$, so you want $$ 0 = h'(c)f(c) + h(c)f'(c) = h'(c)f(c) + h(c)yf(c) = f(c)\Bigl(h'(c)+yh(c)\Bigr).$$ So either $f(c)=0$, or else $h'(c)+yh(c)=0$. You don't want $f(c)=0$ unless $y=0$, so in general you just want $h'(c)+yh(c)=0$.

How about trying to find some function $h(x)$ such that $h'(x) = -yh(x)$ for all $x$? Then use that for $\mathcal{F}$. Obviously, your choice of $h$ will depend on $y$.


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