Question

Let f and g be functions such that f'(X)g(x)≠f(x)g'(x) for all x. Prove that between any roots of f(x) there is a root of g(x).

Answer 1

I believe the following proof is correct. Consider f and g to be functions that are continuously differentiable over the reals. Note that ALL zeros of f (and g) must be isolated (ie, there is no open interval on which f or g are identically zero), as otherwise f' would also be zero on some part of that interval, contradicting the given inequality assumption. Since all zeroes of f are isolated, choose two zeroes, say y and z of f that are consecutive. Consider the function \[h(x)=\frac{f(x)}{g(x)}\] Suppose that h(x) is differentiable everywhere on the interval [y,z]. Then we have that \[h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\] Since f'g-fg' is never zero, this implies that h' is either ALWAYS positive or ALWAYS negative. Suppose h' is always positive, ie that h is increasing on the interval [y,z]. Then since y and z are clearly in the interval, we must have h(z)>h(y), but since f(z)=f(y)=0, this is a contradiction. Thus, our initial assumption that h is differentiable everywhere on [y,z] is false, so h is NOT differentiable everywhere on that interval. Since f and g are known to be differentiable everywhere, this means that the only way differentiability of h can fail on the interval is if g is nonzero somewhere in [y,z]. Thus, g has a zero between y and z. The proof when h' is always negative is similar. Thus, between any two roots of f, g also has a root.

Answer 2

This is great, except for one thing: you put "this means that the only way differentiability of h can fail on the interval is if g is nonzero somewhere in [y,z]." Did you mean that g is zero somewhere in [y,z]?

Answer 3

Yeah, sorry, got caught up in all the different times I had to write "zero" and "nonzero".

Answer 4

It's very good, very nice proof!