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 gf(x) 10 (c) (ii) Solve the equation gf(x) = 2, giving your answer in an exact form. - AQA - A-Level Maths Pure - Question 10 - 2017 - Paper 1

The domain of the inverse function $f^{-1}$ corresponds to the range of the original function $f$. From part (a), we found the range of $f$ to be $\{ y \in \mathbb{R} : y > 4 \}$. Therefore, the domain of $f^{-1}$ is:

$\text{Domain of } f^{-1} = \{ x \in \mathbb{R} : x > 4 \}$

To find the inverse function $f^{-1}(x)$, we start by interchanging $x$ and $y$. We have:

$x = 4 + 3^{-y}$

Rearranging this gives:

$x - 4 = 3^{-y}$

Taking the logarithm on both sides:

$-y \log(3) = \log(x - 4)$

Thus, we have:

$y = -\frac{\log(x - 4)}{\log(3)}$

Consequently, the expression for the inverse function is:

$f^{-1}(x) = -\frac{\log(x - 4)}{\log(3)}$

To find the expression for $gf(x)$, we substitute $f(x)$ into $g(x)$:

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

To solve for $gf(x) = 2$:

$5 - \sqrt{4 + 3^{-x}} = 2$

Rearranging gives:

$\sqrt{4 + 3^{-x}} = 3$

Squaring both sides:

$4 + 3^{-x} = 9$

This simplifies to:

$3^{-x} = 5$

Taking logarithms:

$-x \log(3) = \log(5)$

Thus:

$x = -\frac{\log(5)}{\log(3)}$