Reflecting a point by a function Is it possible to reflect a point $ (x_1,y_1) $ by a function that is continuous everywhere?
For example, reflection of any point by the function $ f(x) = 0$ would be
$$
(x_1,y_1) \Longrightarrow (x_1, -y_1)$$
Is there a way to find where the point is after a reflection by any function? 
 A: This is equivalent to finding the reflection of a point $(x_1,y_1)$ across the tangent line of a function. The tangent line chosen can be arbitrary (see my comment regarding the reflection of $(0,1)$ across $x^2$), so we will express our new point $P$ as a function of $c$, which is the value of $x$ corresponding to some tangent line.

Given a line $y = ax + b$, let us define $$D := \frac{2x_1 + 2a\,(y_1 - b)}{1 + a^2} \tag{1}$$
Then the reflection of a point $(x_1,y_1)$ is $$(D - x_1, aD - y_1 + 2b) \tag{2}$$
Now, the tangent line $L$ of a function $f(x)$ is given by $$L(c) = f'(c)\cdot(x-c)+f(c)$$ or equivalently$$L(c) = f'(c) \, x -c \, f'(c)+f(c) \tag{3}$$
where $f'(c)$ is the derivative of $f$ evaluated at $x=c$.
And since $L$ is a line of the form $ax+b$ (with $a = f'(c)$ and $b = -c \, f'(c)+f(c)$), we substitute $(3)$ into $(1)$ to get
$$\color{green}{D := \frac{2x_1 + 2f'(c)\,\big(y_1 + c \, f'(c)-f(c)\big)}{1 + f'(c)^2}} \tag{4}$$ For convenience, let us also define
$$\color{purple}{H := 2\big(f(c) - c \, f'(c)\big)} \tag{5}$$
Using this definition of $D$, we combine $(2)$, $(4)$, and $(5)$ and conclude that
$$P(x_1,y_1,c) = (\color{green}D - x_1, \,\, \color{green}D\cdot f'(c) - y_1 + \color{purple}H)$$
You might be able to build on this by using $c$ such that the normal at $c$ is the closest line to your point, but you may have issues with symmetrical functions. 

You can see this method in action here.
