# A question on the relations between a function and its derivatives.

Got stuck on this one -

Let $f:\mathbb R \to \mathbb R$ be differential twice. Show that if there are three different solutions to the equation $f(x)^2=x^2$ then there is at least one solution to the equation $f''(x)f(x)+f'(x)^2=1$

Thanks!

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Order the three different solutions and consider the first and second solution. Since $f(x)^2-x^2$ is zero at these points, by the mean value theorem it must have zero derivative at some intermediate point. The same is true for the second and third solution, so we have two different points at which $(f(x)^2-x^2)'=2f'(x)f(x)-2x=0$. Applying the mean value theorem to these two points in turn shows that there is a point at which $(2f'(x)f(x)-2x)'=2f''(x)f(x)+2f'(x)^2-2=0$.
Hint: Take the second derivative of $f(x)^2$ (in general, not as $x^2$).
That seems like a slightly opaque and potentially misleading hint to me -- it might be taken to mean that you just have to differentiate $f(x)^2=x^2$ twice. –  joriki Jan 22 '13 at 17:20
I just wanted to give him the simplest means to understand the equation $f'' f+f^2=1$. That was it. I understand your point, though, and will clarify. –  Ron Gordon Jan 22 '13 at 17:23
Hey, thanks. I did derivate both sides and got $f''(x)f(x)+f'(x)^2=1$. If I say that $g(x)=f(x)^2$, and that g(x) have three solutions, then derivating it twice making it have at least one solution. But why? –  Harold Jan 22 '13 at 17:28
I think what you can say is that the two equations are equivalent, so any function $f$ that satisfies the former equation also satisfies the latter. –  Ron Gordon Jan 22 '13 at 17:46
@rlgordonma: From your further comments, it now seems that you did mean that you just have to differentiate the equation $f(x)^2=x^2$ twice. That's not true; $f$ only satisfies $f(x)^2=x^2$ at three isolated points; to allow us to differentiate the equation, it would have to hold on some interval. Note that if your solution were correct the hypothesis that there are three different solutions wouldn't be needed; one would be enough. –  joriki Jan 22 '13 at 22:33