Stretching a rubber band Question
Is it true that if you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position? Give reasons for your answer.
My attempt at the solution
Let us take a piece of un-stretched rubber band and place it on $x$-axis such that one end is at $x=a$ and the other end is at $x=b$. Therefore, the length of this un-stretched rubber band is $l=b-a$
Let us mark $n$ number of points on the rubber band, with equal gap $d_0$ between each point. Now, the $x$ co-ordinate of $k^{\text{th}}$ point is given by
$$
p(x_k)=x_0+(k-1)d_0
$$
where $x_0$ is the $x$ co-ordinate of the first point, which is $a$ and
$$
d_0=\frac{l}{n-1}=\frac{b-a}{n-1}
$$
Therefore,
$$
p(x_k)=a+(k-1)\left(\frac{b-a}{n-1}\right)
$$
Let us now stretch the rubber band on both ends such that the rubber band now spans between $x=c$ (which is to the left of $a$) and $x=d$ (which is to the right of $b$) on the $x$-axis. Therefore, $c<a$ and $d>b$. Now, the $x$ co-ordinate of $k^{\text{th}}$ point is given by
$$
p_1(x_k)=c+(k-1)\left(\frac{d-c}{n-1}\right)
$$
Let us now assume that some $k^{\text{th}}$ point before and after stretching end up in the same position. Therefore,
$$
p(x_k)=p_1(x_k) \\
a+(k-1)\left(\frac{b-a}{n-1}\right)=c+(k-1)\left(\frac{d-c}{n-1}\right)
$$
Simplifying, we end up with
$$
k=\frac{n(a-c)+(d-b)}{(a-c)+(d-b)}
$$
Notice that $(a-c)$ is the by how much we stretched on the left and $(d-b)$ is the same on the right. Let us assume that we stretched the band unevenly such that
$$
a-c=m(d-b):m>0
$$
Therefore,
$$
k=\frac{n(a-c)+(d-b)}{(a-c)+(d-b)}=\frac{nm+1}{m+1}
$$
Intuitively speaking, if $m=1$, that is, if we stretch by the same magnitude on both ends, then
$$
k=\frac{n+1}{2}
$$
which makes sense because the mid-point on the band would remain at the same place.
Since $k$ has a real value, it could be said that there exists a point which remains at the same place after stretching the band.
The question is asked in a calculus book and I couldn't think of a calculus solution and I came up with this. Is this valid? What would the calculus solution look like?
 A: I don't think you're supposed to assume anything about $p$ other than continuity.  Consider the function $f(x)=p(x)-x$. We have $f(a)=p(a)-a<0$ and $f(b)=p(b)-b>0$.  By the intermediate value theorem, there is a point $a<c<b$ with $f(c)=0$.  Then $c$ is a fixed point of $p$.
A: The horizontal axis indicates the position of a point on the rubber band, and the vertical axis its position "after" transformation.  In the initial state, these positions are the same (of course), and thus described by the equation $y = x$, shown in blue.
After arbitrary stretching, the right-most point of the rubber band is pulled to the right (hence the corresponding point is higher along the $y$ axis).  Conversely the left-most point on the rubber band is pulled to the left (hence the corresponding point is lower along the $y$ axis), as shown in the tan graph.  We must assume the stretching is continuous (no breaks or rips in the rubber band), but otherwise arbitrary.

By simple topology (the intermediate value theorem), there must be a point where the continuous tan graph crosses the blue graph.  At that point, the $y = x$... and corresponds to a point that has not moved.
Note that in complex nonlinear stretching you can have 1, 3, or any odd number of crossings of the two curves, and thus any such number of fixed points.
(You can even have whole segments that match, technically giving and infinite number of such points.  Moreover, the functions can touch but not cross at points... but so long as the left-most point of the tan curve is below the blue, and the right-most point is above the blue, you MUST have at least one crossing point.)
