Reflection of light from function graph Let a positive convex decreasing differentiable function $f(x)$ be defined on $\mathbb{R}$  and $\lim_{x \to +\infty}f(x)=0.$ Let the point light source be placed at $ P(x_0,y_0)$ with $ y_0>0,\,y_0 <f(x_0).$  Light is assumed to be reflected from the plot $y=f(x)$ and the $x$-axis. Does there exist a number $R$ s.t. the part of the graph $y=f(x)$
 for $x>R$ is not lightened? 
The model example $f(x):=e^{-x},\,P(0,0.5)$ suggests the answer is yes.
The question is migrated from SE. 
 A: Just empirically, I believe the OP's $e^{-x}$ example has the property that 
ray reflections quickly become increasingly vertical,
and so will not reach arbitrarily large $x$:

         


(I did not, however, perform the calculations to prove this.)
A: In the paper The existence of unbounded oscillating trajectories in a problem of billiards (1962) Leontovich proved that under bell-like curve (it must be zero at  $\pm\infty$) each trajectory oscillates, i.e. it crosses y-axis infinitely often. Also he proved that among all trajectories do exist finite and infinite ones.
A: If a ray of light at angle $\alpha$ above the horizontal hits your curve $y = f(x)$ from below at a point where the tangent to the curve has angle $\beta$ below the horizontal, it will reflect at angle $\alpha + 2 \beta$ below the horizontal, and then come back up at $\alpha + 2 \beta$ above the horizontal.
In particular, if $\alpha + 2 \beta = \pi/2$ it goes vertically down (and then retraces itself backwards), and if  $\alpha + 2 \beta > \pi/2$ it goes backwards (i.e. to the left). 
Let the $n$'th reflection on the curve take place at $(x_n, y_n)$, with incoming ray at angle $\alpha_n$.  Then we have
$$\eqalign{\alpha_{n+1} &= \alpha_n - 2 \arctan(f'(x_n))\cr
           y_{n+1} + y_n &= \tan(\alpha_{n+1}) (x_{n+1} - x_n)\cr
           y_{n+1} &= f(x_{n+1})}$$
Thus $$\dfrac{\Delta \alpha_n}{\Delta x_n} = \dfrac{\alpha_{n+1}-\alpha_n}{x_{n+1} - x_n} = \tan(\alpha_{n+1}) \dfrac{- 2 \arctan(f'(x_n))}{
f(x_{n+1}) + f(x_n)} $$
In order for $x_n \to \infty$ with $\alpha_n$ increasing but staying below $\pi/2$, we would certainly need this to go to $0$.  In the case $f(x) = e^{-x}$, that certainly won't happen, as $\arctan(f'(x_n)) \approx f'(x_n) = - f(x_n)$, while
$f(x_{n+1}) + f(x_n) < 2 f(x_n)$.  More likely candidates would be functions
$f$ that go to $0$ very slowly, perhaps something like $1/\log(x)$.
