Rolle's Theorem Problem with f(a)=f(b)=0: Finding f'(c)=f(c)/c on [a,b]

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Anyway, I've come across this problem that I can't figure out. It looks set up as a Rolle's Thm problem, but it just doesn't work out...

Let f be a continuous function on [a,b] and differentiable on (a,b) for some a,b > 0. Suppose f(a) = f(b) = 0.

Show that f'(c)=f(c)/c for some c between a and b.

The thing is this is next to straightforward problems like the integral of xsinx and |3-2x|>1. Am I just missing something?

I mean, I've spent about 3 hours looking for a similar problem/theorem in my Calc book; even just a pointer from someone who knows the answer would be great.
 
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Use Rolle's theorem on g(x) = f(x)/x
 
*smacks self hard*

I can't believe I missed that, thanks.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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