The functions f and g are defined by $f: x \mapsto 3x + \ln x, \; x > 0, \; x \in \mathbb{R}$ $g: x \mapsto e^x, \; x \in \mathbb{R}$ (a) Write down the range of g - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

To find the composite function $fg$, we substitute $g(x)$ into $f(x)$:

$fg(x) = f(g(x)) = f(e^x) = 3e^x + \ln(e^x) = 3e^x + x.$

Thus, the composite function is defined as:
$fg: x \mapsto -x^2 + 3e^x, \; x \in \mathbb{R}.$

The term $3e^x$ dominates the behavior of the function $fg(x) = -x^2 + 3e^x$. Since $-x^2$ is a downward-opening parabola with a maximum point, and $3e^x$ grows to infinity, the range of $fg(x)$ is $[3, \infty)$.

To solve the equation $\frac{d}{dx}[g(f(x))] = x(xe^x + 2),$ we first compute the left-hand side.

Using the chain rule: $g(f(x)) = g(3x + \ln x) = e^{3x + \ln x} = e^{3x}x,$
thus, $\frac{d}{dx}[g(f(x))] = \frac{d}{dx}[e^{3x}x] = 3e^{3x}x + e^{3x} = e^{3x}(3x + 1).$
Now setting the equation:

$e^{3x}(3x + 1) = x(xe^x + 2).$

To find solutions, we get: $e^{3x}(3x + 1) \ -\ x^2 e^x - 2x = 0.$
This will lead to further simplification and solving iteratively. The exact solutions require numerical methods or specific values. Some obvious solutions are $x=0$ and checks on other intervals show no others might exist. Finally, evaluate $e^{3x} \neq 0$ leads us to:

$x = 0, \; 6.$