Cauchy's mean value theorem conditions

Cauchy's mean value theorem conditions I have seen some statements of the Cauchy's mean value theorem that requires g′ is never 0 on (a,b), and some others that require f′ and g′ to never be simultaneously zero.

Lemma: If functions f and g are both continuous on the closed interval [a,b], and differentiable on the open interval (a,b) and g′ is never zero on (a,b), then there exists some c∈(a,b), s.t f′(c)/g′(c)=f(b)−f(a)/(g(b)−g(a)).

Why is it needed for g′ to never be 0 on all (a,b) , don't we just need g′ not to be zero on that value c ?

Solution 1:

Cauchy's MVT with the fewest restrictions is the following:

Theorem 1.

Suppose $f,g:[a,b]\to\Bbb{R}$ are functions which are continuous on $[a,b]$, and differentiable on $(a,b)$. Then, there is a point $c\in (a,b)$ such that $f'(c)[g(b)-g(a)]=g'(c)[f(b)-f(a)]$.

The proof is a one-liner, by applying Rolle's theorem to the function $h:[a,b]\to\Bbb{R}$ defined as \begin{align} h(x)=f(x)[g(b)-g(a)]-g(x)[f(b)-f(a)]. \end{align}

When written out in this multiplicative form, there is no issue at all. If however you wish to write things in the fraction form, one has to ensure that the denominators are non-zero. So, we need to know that $g(b)-g(a)\neq 0$ , and also that $g'(c)\neq 0$. Then, we can write the conclusion as \begin{align} \frac{f'(c)}{g'(c)}&=\frac{f(b)-f(a)}{g(b)-g(a)}. \end{align}

You ask

Why is it needed for g′ to never be 0 on all (a,b) , don't we just need g′ not to be zero on that value c ?

well, the answer is that the theorem is an existence result, so you have no way of determining a-priori whether or not the particular $c$ found from the theorem is such that $g'(c)\neq 0$. This is why sometimes one adds extra hypotheses to ensure that the derivatives are never zero:

Theorem 2.

If $f,g:[a,b]\to\Bbb{R}$ are continuous on $[a,b]$ and differentiable on $(a,b)$, and $g'$ is nowhere vanishing on $(a,b)$, then $g(b)\neq g(a)$ and there is some $c\in (a,b)$ such that $\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$.

The proof is in two stages. First apply theorem 1 to find some point $c$. Next, $g'$ nowhere vanishing implies by Darboux's theorem that $g'>0$ on all of $(a,b)$ or that $g'<0$ on all of $(a,b)$. This implies either that $g$ is strictly increasing or strictly decreasing, and thus $g(b)\neq g(a)$. Therefore, we can make the appropriate divisions in theorem 1.

Once again, the extra hypothesis $g'$ non-vanishing on $(a,b)$ is simply there to ensure that $g(b)\neq g(a)$, and so that no matter the point $c$ which we find (because note that there could be many values of $c$ which make the equation in theorem 1 true), we are guaranteed that $g'(c)\neq 0$.

If you want to be super explicit, consider the following example. Let $[a,b]=[-1,2]$, and let $f=0$ be the zero function. Let $g(x)=x^2$. Then, in theorem 1, any value of $c\in [-1,1]$ will make the equation true, simply because both sides are trivially equal to zero for all values of $c$. So, in particular, the equation is true for $c_1=0$: \begin{align} f'(0)[g(2)-g(-1)]&=g'(0)[f(2)-f(-1)] \end{align} But we cannot write this in the divided form because $g'(c_1)=g'(0)=0$. On the other hand, for $c_2=1$, the conclusion of theorem 1 is still true: \begin{align} f'(c_2)[g(2)-g(-1)]&=g'(c_2)[f(2)-f(-1)]=0. \end{align} In this case, we can write things in the divided form: $\frac{f'(c_2)}{g'(c_2)}=\frac{f(2)-f(-1)}{g(2)-g(-1)}$ (again both the numerators are zero, and denominators are non-zero).

I hope you see the issue: in the abstract statement of the theorem, all we know is that there is some value of $c$, and there could be multiple of them, as I've shown in this example. Here, it turns out that one of the choices of $c$, namely $c_1=0$, is "bad" for the division form, but the other choice $c_2=1$ is perfectly fine. But in writing down a general theorem, you can't guarantee beforehand which is going to happen. Therefore, if you want to write the conclusion of Cauchy's MVT in the divided form, you should make the extra hypothesis that $g'$ is non-vanishing.

Anyway, long story short, the multiplied version is so simple that I suggest you remember that.