The function f is defined by $f: x \mapsto 4 - \ln(x + 2), \quad x \in \mathbb{R}, \quad x > -1$ (a) Find $f^{-1}(x)$ - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 3

The domain of $f^{-1}$ corresponds to the range of the original function $f$. Since $f(x) = 4 - \ln(x + 2)$ and its output can be seen from behavior:

• As $x \to -1^{+}$, $f(x) \to 4 - \ln(1) = 4 - 0 = 4$.
• As $x \to \infty$, $f(x) \to -\infty$.

Therefore, the range of $f$ is $(-\infty, 4]$, which gives us the domain of $f^{-1}$ as:

$x \leq 4$

To find $fg(x)$, we first substitute $g(x)$ into $f$:

$fg(x) = f(g(x)) = f(e^{x} - 2)$

Using the definition of $f$:

$= 4 - \ln((e^{x} - 2) + 2)$

This simplifies to:

$= 4 - \ln(e^{x})$

which can further be simplified as:

$= 4 - x$

From the previous answer, we have:

$fg(x) = 4 - x$

The output will vary as $x$ takes any real value in $ext{R}$. As $x \to -\infty$, $fg(x) \to +\infty$, and as $x \to +\infty$, $fg(x) \to -\infty$. Hence, the range of $fg$ is:

$fg(x) \leq 4$