The function f is defined by f: x ↦ 4 − ln(x + 2), x ∈ ℝ, x > −1 (a) Find f^{-1}(x) - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 3

The function f(x) maps x > -1 into a range. From the function, we see the maximum value occurs when x approaches 0:

When x approaches 0, f(0) = 4 - ext{ln}(2).

As x approaches ∞, f(x) approaches 4.

Thus, the domain of f^{-1}(x) is:

eq 4$$; However, since the range will be (−∞, 4) based on the properties of the natural logarithm, we can express it as: $$x ext{ ≤ } 4$$.

To find fg(x), we start with:

$fg(x) = f(g(x))$

Substituting g(x):

$g(x) = e^{-x} − 2$

Therefore:

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

Substituting into f gives:

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

This simplifies to:

$fg(x) = 4 - ext{ln}(e^{-x}) = 4 + x$

So, in simplest form:

$fg(x) = 4 + x$.

The range of fg(x) = 4 + x depends on the input x.

As x can be any real number (x ∈ ℝ), the output also covers all real numbers:

Thus, the range of fg is:

$fg(x) ≤ 4$.