The functions f and g are defined by $f: x ightarrow 3x + ext{ln}x, ext{ for } x > 0, ext{ where } x ext{ is in } eals$ g: x ightarrow e^{x}, ext{ where } x ext{ is in } eals$ (a) Write down the range of g - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 2

The composite function $fg$ is given by:

$fg(x) = g(f(x)) = g(3x + ext{ln}x) = e^{(3x + ext{ln}x)} = e^{3x} imes e^{ ext{ln}x} = e^{3x} imes x.$

Thus,

eals.$$

To determine the range of $fg(x) = x imes e^{3x}$, note that for $x > 0$, the product of a positive number $x$ and the exponential function $e^{3x}$ (which is always positive) results in positive values.

For $x < 0$, although $x$ is negative, $e^{3x}$ approaches 0 as $x$ decreases, causing $x imes e^{3x}$ to approach 0 from the negative side but never going above 0.

Thus, the range of $fg$ can be expressed as:

ightarrow (- ext{∞}, ext{ } 0) ext{ and } fg(x) ightarrow (0, ext{ } + ext{ } ext{∞}).$$

First, find the derivative of the composite function:

Using the product rule: dx(g(f(x))) = e^{3x}(3x + 1) + x imes 3 e^{3x} = e^{3x}(3x + 1 + 3x) = e^{3x}(6x + 1).$$ Setting this equal to the right-hand side of the equation: e^{3x}(6x + 1) = x(x e^{2} + 2). Rearranging this gives: $$6x + 1 = rac{x(x e^{2} + 2)}{e^{3x}}. Two cases appear: 1. $e^{3x} eq 0$, process the equation further. 2. If $e^{3x}$ approaches $0$, $e^{3x} eq 0$: $$x = 0, 6$$ (only integer solutions).