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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.

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    @Ross Millikan, @JimJones: almost certainly, but it was like this in the original; I only LaTeX-ed it. I'll fix that.2010-11-23

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$\textbf{HINT:}$

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)

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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 $y$.