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 6 - 2009 - Paper 2

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

1. First, compute $g(x) = e^x$.
2. Then, substitute into $f$: $fg(x) = f(g(x)) = f(e^x) = 3e^x + \ln(e^x).$
3. Simplifying this gives: $fg(x) = 3e^x + x.$ Thus, we have: $fg: x \mapsto -x^2 + 3e^x, \; x \in \mathbb{R}$.

To determine the range of $fg(x) = -x^2 + 3e^x$, we need to analyze the behavior of the function. As $x \to -\infty$, $e^x \to 0$, and the dominating term is $-x^2$, so $fg(x) \to -\infty$. As $x \to \infty$, the exponential growth of $3e^x$ dominates the quadratic term, hence $fg(x) \to \infty$. Therefore, the minimum value occurs at local extrema, which can be found through differentiation. Ultimately, the range of $fg$ is:

$fg(x) \geq 3.$

To solve the equation, we start by finding the derivative of the composite function:

1. Calculate: $\frac{d}{dx}[g(f(x))] = \frac{d}{dx}[e^{f(x)}] = e^{f(x)} \cdot f'(x).$
2. For $f(x) = 3x + \ln x$, the derivative is: $f'(x) = 3 + \frac{1}{x}.$
3. Therefore, we have: $e^{3x + \ln x} \cdot \left(3 + \frac{1}{x}\right) = xe^x (xe^x + 2).$
4. Simplifying this equation allows us to solve for x: $$$$

ewline2x + 6x^2 e^x = x e^x (x + 2),$leading to:$6x - 6e^x = 0 \Rightarrow x = 0, 6.$$