I think I am misunderstanding the Cauchy Mean Value Theorem Right before I posted this question, I arrived at an answer (which I have posted below). I figured I would publish this question anyways in case other people had similar confusions regarding the Cauchy Mean Value Theorem.

The Cauchy Mean Value Theorem reads as follows:

If $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, then there is a number $x$ in $(a,b)$ such that: \begin{align} [f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x) \end{align} Further, assuming $g(b)\neq g(a)$ and $g'(x)\neq 0$, we may write: \begin{align}\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(x)}{g'(x)} \end{align}

Here is a hand-drawn graph depicting a specifically chosen $f$ and $g$ with the following properties:

*

*$f(a)=g(a)$ and $f(b)=g(b)$


*$g(b)-g(a)=f(b)-f(a) \lt 0$


*$f$ is strictly increasing on the interval $[a,d]$ and $f$ is strictly decreasing on the interval $[d,b]$


*$g$ is strictly decreasing on an interval $[a,c]$ and $g$ is strictly increasing on the interval $[c,b]$.


*$c \lt d$


*From the drawing, we conclude that $f$ and $g$ are continuous and differentiable throughout the interval

Given how I have defined these functions, I feel as though it is impossible for such an $x$ to exist.
By the Mean Value Theorem, we know there is some $y_g$ such that $g(b)-g(a)=g'(y_g)$.
$g(b)-g(a)$ is negative, which means $g'(y_g)$ must be negative. Such a $y_g$ is restricted to somewhere within the strictly decreasing interval $[a,c]$; a negative slope cannot be found on a strictly increasing interval, right?
A repeated application of the Mean Value Theorem to the function $f$ along the interval $[a,b]$ will show that there is some $y_f$ such that $f(b)-f(a)=f'(y_f)$. Clearly, $f'(y_f)$ is negative. A negative slope cannot be found on a strictly increasing interval, so $y_f$ must be somewhere in $[d,b]$. Notice that because $c\lt d$, we must have that $[a,c] \cap [d,b] = \emptyset$.
So which assumption have I violated to yield this contradictory result? Alternatively, where is the lapse in my logic?

 A: The reason this approach is incorrect is because it incorrectly assumes that the $x$ that is posited to exist that solves the equation $[f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x)$ is always an object that simultaneously solves the following two equations:
$\frac{f(b)-f(a)}{b-a}=f'(x)$
$\frac{g(b)-g(a)}{b-a}=g'(x)$
Consider the following two logical statements:
$\left(\frac{f(b)-f(a)}{b-a}=f'(x) \text{ and } \frac{g(b)-g(a)}{b-a}=g'(x) \right) \implies [f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x)$
$[f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x) \implies \left(\frac{f(b)-f(a)}{b-a}=f'(x) \text{ and } \frac{g(b)-g(a)}{b-a}=g'(x) \right)$
The first statement is always true. The second statement is, in general, false.
Only under specific circumstances is the second statement true. Rearranging the equation and throwing $b-a$ into the mix, you can produce:
$\frac{f(b)-f(a)}{b-a}=f'(x)\cdot \left[\frac{g(b)-g(a)}{g'(x)\cdot(b-a)} \right]$ and $\frac{g(b)-g(a)}{b-a}=g'(x)\cdot \left[\frac{f(b)-f(a)}{f'(x)\cdot(b-a)} \right]$. From this, you can deduce what those "special circumstances" are.
The take home picture is that the $x$ we are looking for does not have to be the $x$ that is involved in each function's individual Mean Value Theorem.
