Composition and Addition on a Set of Function and Consequences for Distributitivity I'm working on the following question:

Let $S$ be the set of all functions $f: \mathbb{R} \rightarrow \mathbb{R}$. Consider the following pointwise operations $*$ and $+$:
  $$(f+g)(x) = f(x) + g(x) \\ (f*g))(x) = f(g(x))$$
  Which of the following is true:
  
  
*
  
*$f * (g + h) = f*g + f*h$
  
*$(g + h)*f = g*f + h*f$

By using the example of constant functions one can eliminate 1. leaving 2. But, I only know this because I peeked at the answer- I didn't think of this beforehand.
Is there a way of seeing this directly without examples? For instance $S$ appears to have some algebraic structure.
 A: At least the first one is false. Take $f(x) = x^2 $, $g=x$ and $h=1$, then 
$$ f * (g+h) = f(x+1) = (x+1)^2 $$
but
$$ f * g = x^2 $$
$$ f * h = 1 $$
A: For the second one apply the definitions: Let's take some $x\in\mathbb{R}$ and look at $((g+h)*f)(x)$. By the definition of $*$, this is the same as $(g+h)(f(x))$, which we can further simplify using the definition of $+$ to $(g(f(x))+h(f(x))$. But this is nothing else than $(g*f)(x)+(h*f)(x)$, so $(g+h)*f=g*f+h*f$.
A: For (1): It is saying given $f,g,h$ can we have 
$$f(g+h)=f(g)+f(h).$$
It almost reminds you of linearity. So any function $f$ which is not behaving "linearly" (so to speak) can provide a counter example. For example, take $f(x)=\sin x$ and take $g(x)=h(x)=x$. Then if this was true, we would have $\sin(2x)=2\sin x$, which as you may know is not true for all $x$.
For (2). This one is true. Here we need to show that $(g+h)f=g(f)+h(f)$. Let $a \in \mathbb{R}$, then 
\begin{align*}
(g+h)f(a) & = g(f(a))+h(f(a)) && (\text{by definition})\\
& = g(f)(a)+h(f)(a).
\end{align*}
So the two functions agree at all real numbers, hence they are equal functions.
