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Let $f(x)$, $g(x)$ be two differentiable functions, such that for every $x$, f'(x)g(x)≠ f(x)g'(x) Also, $f(a) = f(b) = 0$. Prove that there exists $c\in[a,b]$ such that $g(c) = 0$.


The first thing I did was to isolate the inequality into something more intuitive, and so I got:

\frac {f'(x)}{g'(x)} ≠ \frac{f(x)}{g(x)} Meaning that the ratio of the slopes of the two functions can never be the same as the ratio of the y-values.

The information that $f(a)=f(b)=0$ tells me, according to Rolle's theorem, that there exists a value $c\in[a,b]$ such that f'(c) = 0.

I also know that according to the Cauchy Mean Value theorem, $\exists c\in[a,b]$ such that f'(c)[g(b)-g(a)] = g'(c)[f(b)-f(a)] So I know two things about the ratio of the slopes.

I also know that $\frac {f(x)}{g(x)} ≠ \frac {f(b)-f(a)}{g(b)-g(a)}$ and since $f(b) = f(a)$, I have $\frac {f(x)[g(b)-g(a)]}{g(x)} ≠ 0$ .

I don't really know how to continue from here, or if anything i've done so far is even in the right direction. $g(x)$ is in the denominator so how can I prove that it can equal $0$ for a value between $a$ and $b$ ?

Please limit answers to the scope of what I'm supposed to know taking a first-semester calculus course.

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Assume that $g(x)\ne0$ for every $x$ in $[a,b]$. Then the function $h:x\mapsto f(x)/g(x)$ is well defined, $h(a)=h(b)=0$, and h'(x)\ne0 for every $x$ in $[a,b]$. This contradicts a theorem you know.