Given f(x) = (2x+1)/(3x+3) and g(x) = 3x+2. Find g(f(x)) and the invariant elements of g(x).

Given $f(x) = \frac{2x+1}{3x+3}$ and $g(x) = 3x+2$.

$$g(f(x)) = g\left(\frac{2x+1}{3x+3}\right) = 3\left(\frac{2x+1}{3x+3}\right) + 2$$

$$g(f(x)) = \frac{3(2x+1)}{3(x+1)} + 2$$ $$g(f(x)) = \frac{2x+1}{x+1} + 2$$

To combine the terms, find a common denominator:

$$g(f(x)) = \frac{2x+1}{x+1} + \frac{2(x+1)}{x+1}$$ $$g(f(x)) = \frac{2x+1 + 2x+2}{x+1}$$ $$g(f(x)) = \frac{4x+3}{x+1}$$

The domain of $g(f(x))$ requires $x$ to be in the domain of $f$, and $f(x)$ to be in the domain of $g$.
The domain of $f(x)$ is $x \in \mathbb{R}, x \neq -1$. The domain of $g(x)$ is $x \in \mathbb{R}$.
Since $f(x)$ will always produce a real number (for $x \neq -1$), and $g(x)$ accepts all real numbers, the only restriction is $x \neq -1$.
Therefore, the domain of $g(f(x))$ is $x \in \mathbb{R}, x \neq -1$.

The function $g(f(x))$ expressed in a similar manner is:

$$\boxed{g(f(x)): x \mapsto \frac{4x+3}{x+1}, x \in \mathbb{R}, x \neq -1}$$

Given $g(x) = 3x+2$.
We need to solve $g(x) = x$:

$$3x+2 = x$$

Subtract $x$ from both sides:

$$3x - x + 2 = 0$$ $$2x + 2 = 0$$

Subtract 2 from both sides:

$$2x = -2$$

Divide by 2:

$$x = -1$$

The domain of $g$ is $x \in \mathbb{R}$. Since $-1$ is a real number, it is in the domain of $g$.

The element in the domain of $g$ which is invariant under $g$ is:

$$\boxed{x = -1}$$