Darboux's Theorem. Let $I$ be an open interval, and let $f : I \to R$ be a differentiable function. If $a$ and $b$ are points of $I$ with $a < b$ and if $y$ lies between $f' (a)$ and $f' (b)$, then there exists a number $x\in [a, b]$ such that $f'(x) = y$.
We will prove the following equivalent version of the above theorem,
Equivalent Version of Darboux's Theorem. Let $I$ be an open interval, and let $f : I \to R$ be a differentiable function. If $a$ and $b$ are points of $I$ with $a < b$ such that $f' (a)<0$ and $f' (b)>0$, then there exists a number $x\in (a, b)$ such that $f'(x) = 0$.
Proof
$\color{red}{\text{Observation 1.}}$ First of all let us note that the above version of Darboux's Theorem (for the sake of completeness we should prove equivalence but I am skipping it for now) is true if the restriction of $f$ to $[a,b]$ is not injective on $[a,b]$. To see why this claim is true, let us note that if $f$ is not injective on $[a,b]$ then there exists some $c,d\in[a,b]$ such that $f(c)=f(d)$ but $c<d$. So by Rolle's Theorem we can say that, $f'(x)=0$ for some $x\in(c,d)$ and hence for some $x\in(a,b)$. Hence in this case we are done.
$\color{red}{\text{Observation 2.}}$ So we can now say that if we can show that $f$ can't be injective on $[a,b]$ then we are done. Suppose that $f$ is injective on $[a,b]$. Now notice that $f$ is a continuous and injective mapping. So, by this result $f$ must be strictly monotone on $[a,b]$. Without loss of generality let's assume that it's strictly increasing.
So, $a<x$ will imply that $f(a)<f(x)$. Consequently, $\dfrac{f(x)-f(a)}{x-a}>0$ for all $x>a$. It follows that, $$\displaystyle\lim_{x\to a^+}\dfrac{f(x)-f(a)}{x-a}\ge0$$But this is a contradiction since $f'(a)<0$. So, $f$ can't be injective.
The case when $f'(a)>0$ and $f'(b)<0$ can be shown in a similar manner or by considering the function $g(x)=-f(x)$ and applying the above argument.
Is there any flaw in my argument?
