I have the following question before me: $f$ and $g$ are two functions derivable in $(a,b)$ such that $f(x) g'(x)-f'(x)g(x)>0$ for all $x$ in $(a,b)$ then prove that there lies exactly one root of $g(x)=0$ in between two consecutive roots of $f(x)=0$.
I assumed the existence of two roots of $g(x)=0$ in between two consecutive roots of $f(x)=0$ in order to reach a contradiction. Here is how I attempted: Let $c$ and $d$ be the two consecutive roots of $f(x)=0$. Now for all $x$ in between $c$ and $d$, $f(x)$ is non-zero. Consider the function $g(x)/f(x)$ in the interval $(c,d)$. The derivative of this function is $ (f(x)g'(x)-g(x)f'(x))/((f(x))^2$.
Due to the given condition, this derivative is always positive which implies that the function is strictly increasing in nature and thus can't have equal values at any two different points in its domain. Now I consider the existence of two consecutive roots of $g(x)=0$ in $(c,d)$ which makes the value of the function $g(x)/f(x)$ $0$ at two different points which is impermissible. Thus I am able to arrive at a contradiction. From here I can conclude that there lies at most one root of $g(x)=0$ in between two consecutive roots of $f(x)=0$.
But the question asks me to show the existence of exactly one such root. How do I proceed from here to prove this? Please suggest.
Assume that $g$ has no zeros in between $c$ and $d$. Consider the function $f/g$ . It has two roots at $c$,$d$ thus it's derivative has a root in $(c,d)$ but you can see that derivative of $f/g$ is always negative. Contradiction so no. of roots is greater or equal to 1 and by your previous argument no. of roots is exactly 1.
Or you can argue as follows Since its derivative is negative so $f/g$ is strictly decreasing and now $f/g$ has a zero at $c$ then how can $f/g$ have a zero at $d$ ? as it has to increase somewhere to come back to 0.