The intermediate value theorem

I consider in this entry the exercise 2 of the homework page and some people asked. The interest of this exercise is to show what we gain in this course comparing we know from the high-school or Calculus I. We learnt the following intermediate value theorem: if $f$f is a smooth function in $[a,b]$, then there exists  $\xi\in (a,b)$ such that $f(b)-f(a)=f'(\xi)(b-a)$. If we write this equality as  $$f'(\xi)=\frac{f(b)-f(a)}{b-a}$$ the results asserts that there exists an intermediate point between $a$ and $b$ where the tangent line at $x=\xi$ is parallel to the straight-line joining the points $(a,f(a))$ and $(b,f(b))$: the about identity says that the slopes of both lines coincide.

Exercise 1 says that this result holds for any planar curve without to be a graph of a function $y=f(x)$, as it shows the next picture.

A possible starting point of the proof would be applying the intermediate value theorem to the functions  $x=x(t)$ and $y=y(t)$, where $\alpha(t)=(x(t),y(t))$. Suppose $t\in [a,b]$. Then there exists $\xi_1,\xi_2\in (a,b)$ such that $$x(b)-x(a)=x'(\xi_1)(b-a), \ y(b)-y(a)=y'(\xi)(b-a).$$ Hence $$x'(\xi_1)=\frac{x(b)-x(a)}{b-a},\ y'(\xi_2)=\frac{y(b)-y(a)}{b-a}.$$
A vector determining the line $\alpha(a)$ with $\alpha(b)$ is $$\alpha(b)-\alpha(a)=(x(b)-x(a),y(b)-y(a))$$ and a vector of the tangent line at $t$ is $(x'(t),y'(t))$. And we want to prove that the first one is proportional to the second one at some point $t_0\in (a,b)$. But this point is not  $\xi_1$ or $\xi_2$ because they do not need to coincide!

Thus the proof follows other ideas. In the picture we observe that the point that we are looking for is the farthest point of the trace  $\alpha(a,b)$ with respect to the line $R$ joining  $\alpha(a)$ with $\alpha(b)$. Then define $$f:[a,b]\rightarrow\mathbb{R},\ f(s)=\langle \alpha(s)-\alpha(a),N(a)\rangle$$ the distance between $\alpha(s)$ and $R$ and take $s_0\in (a,b)$ the maximum of $f$ (if $\alpha(s_0)=0$, then $\alpha([a,b])\subset R$ and the result is immediate). Let us point out that the maximum is attained because $[a,b]$ is a compact set. In particular, $f'(s_0)=0$ and this yields $$0=f'(s_0)=\langle \alpha'(s_0),N(a)\rangle,$$ which proves the result.

In fact the reasoning says that it holds for any critical point of $f$ in $(a,b)$, as it appears in the next figure.