Spivak, Calculus 3rd Ed. Chapter 11 problem 32

Question

I made it this far through the book but now I'm really stumped. Here's the problem:

Suppose that $f$ and $g$ are two differentiable functions which satisfy $fg'-f'g=0$. Prove that if $a$ and $b$ are adjacent zeros of $f$, and $g(a)$ and $g(b)$ are not both $0$, then $g(x)=0$ for some $x$ between $a$ and $b$. (Naturally the same result holds with $f$ and $g$ interchanged; thus, the zeros of $f$ and $g$ separate each other.) Hint: Derive a contradiction from the assumption that $g(x)\neq 0$ for all $x$ between $a$ and $b$: if a number is not $0$, there is a natural thing to do with it.

The main topics of this chapter are Rolle's theorem, the mean value theorem, the Cauchy mean value theorem, and L'Hopital's rule. Of course by Rolle's theorem $f'=0$ somwhere in the interior, and we can determine that $g'=0$ at the same points. Dividing by $g$ and/or $f$ as suggested hasn't gotten me anywhere.

Can anyone find the contradiction Spivak is talking about?

Answer 1

If $g$ is nonzero, we can divide the given expression by $g^2$, which gives us $\left(\frac{f}{g}\right)'=0$, so $f/g$ is constant on the interval $(a,b)$. Since one of $g(a)$ or $g(b)$ is nonzero one of $(f/g)(a)$ or $(f/g)(b)$ is defined and zero, so we conclude $f/g$ is identically zero on $(a,b)$, so $f$ is zero on $(a,b)$, but then $a$ and $b$ were not adjacent zeros.

Answer 2

Suppose $f(a)=f(b)=0, a<b$ and assume that $f$ does not vanish between $a$ and $b$. [This is the meaning of $a$ and $b$ being adjacent zeros]. If possible, let $g(x) \neq 0$ in $(a,b)$. Note that $(\frac f g)'=\frac {gf'-fg'} {g^{2}}=0$ so $\frac f g$ is a constant $c$ on the interval $(a,b)$. Since $f$ does not vanish between $a$ and $b$ the constant $c$ cannot be $0$. Since the functions are continuous we must have $g=\frac 1 cf$ on the closed interval $[a,b]$. But this equation does not hold when $x=a$ because $f(a)=0$ and $g(a) \neq 0$. This contradiction completes the proof.