If $g(x) = x^2 + 3x + 1$ and $f \circ g = g\circ f$, then $f$ and $g$ intersect on the line $y = x$. 
Let $f:\mathbb R\to \mathbb R$, $g(x) = x^2 + 3x + 1$ and $f∘g=g∘f$.
Prove that $Cf, Cg$ and the line $y=x$ have at least one common point.

My solution:
$fog=gof$ then $f=g$, $Df = Dg = \mathbb R$ and $f(x) = g(x)$
I fear that the above assumption is wrong..
But if it is correct, here is the rest of the solution.
Then $f(x) = g(x) = x^2 + 3x + 1$
Let $f(x)=y$ then $x^2 + 3x + 1 = y$
for $x=y$ we get: $x^2 + 3x + 1 = x$
$x^2 + 2x + 1 = 0$
$(x+1)^2 = 0$
$x = -1$
Because $x=y$ the common point is $(-1,-1)$

Notice that the graphs of $g(x) = x^2 + 3x + 1$ and $y = x$ intersect at precisely one point: $(-1, -1)$. In other words, we know that:
$$\boxed{
g(x) = x \iff x = -1
} \tag{$\star$}$$
To obtain our common point, it is enough to show that $f(-1) = -1$.
To this end, let $k = f(-1)$ and suppose that $f \circ g = g\circ f$. Then in particular, we know that both functions agree at $x = -1$ so that:
$$
g(k) = g(f(-1)) = f(g(-1)) = f(-1) = k
$$
Hence, by $(\star)$, we conclude that $f(-1) = k = -1$, as desired. $~~\blacksquare$
