The function $f$ is defined by $$f(x) = 4 + 3^{-x}, \, x \in \mathbb{R}$$ 10 (a) Using set notation, state the range of $f$ 10 (b) (i) Using set notation, state the domain of $f^{-1}$ 10 (b) (ii) Find an expression for $f^{-1}(x)$ 10 (c) The function $g$ is defined by g(x) = 5 - \sqrt{x}, \, (x \in \mathbb{R} : x > 0) 10 (c) (i) Find an expression for $g f(x)$ 10 (c) (ii) Solve the equation $g f(x) = 2$, giving your answer in an exact form. - AQA - A-Level Maths Pure - Question 10 - 2017 - Paper 1

Since the range of $f$ is $y > 4$, the domain of the inverse function $f^{-1}$ will be:

$f^{-1}(x) : x > 4, \, x \in \mathbb{R}$

To find the inverse function, we start with:

$y = 4 + 3^{-x}$ Interchanging $x$ and $y$ gives: $x = 4 + 3^{-y}$ Subtracting 4 from both sides yields: $x - 4 = 3^{-y}$ Taking the logarithm gives: $-y = \log_3(x - 4)$ Thus, we can express $f^{-1}(x)$ as: $f^{-1}(x) = -\log_3(x - 4)$

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

$g(f(x)) = 5 - \sqrt{4 + 3^{-x}}$

Setting the expression equal to 2, we have:

$5 - \sqrt{4 + 3^{-x}} = 2$ Subtracting 5 from both sides yields: $-\sqrt{4 + 3^{-x}} = -3$ Multiplying through by -1 gives: $\sqrt{4 + 3^{-x}} = 3$ Squaring both sides results in: $4 + 3^{-x} = 9$ Thus: $3^{-x} = 5$ Taking logarithm base 3 leads to: $-x = \log_3(5)$ Therefore: $x = -\log_3(5)$