The functions f and g are defined by f : x ↦ 2|x| + 3, x ∈ ℝ, g : x ↦ 3 - 4x, x ∈ ℝ - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 8

To find fg(1), we need to evaluate g(1) first:

1. Calculate g(1): $g(1) = 3 - 4(1) = 3 - 4 = -1$

2. Now substitute g(1) into f: $f(g(1)) = f(-1) = 2|-1| + 3 = 2(1) + 3 = 2 + 3 = 5$

Thus, fg(1) = 5.

To find the inverse function g⁻¹, we start with the equation of g:

1. Set y = g(x): $y = 3 - 4x$

2. Rearranging to find x in terms of y: $4x = 3 - y$ $x = \frac{3 - y}{4}$

Thus, the inverse function is: $g^{-1}(y) = \frac{3 - y}{4}$

In terms of x, we can write: $g^{-1}(x) = \frac{3 - x}{4}$

We start by letting g(x) = z, therefore: $g(g(x)) + [g(x)]^2 = 0$ => (g(z) + z^2 = 0) $g(z) = 3 - 4z$ $3 - 4z + z^2 = 0$ Rearranging gives: $z^2 - 4z + 3 = 0$ Applying the quadratic formula: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}$ Therefore, z can be: $z = 3 ext{ or } z = 1$

Recalling that z represents g(x), we now solve for x:

1. For z = 3: $3 - 4x = 3 \implies x = 0$

2. For z = 1: $3 - 4x = 1 \implies 4x = 2 \implies x = \frac{1}{2}$

Thus, the solutions are x = 0 and x = \frac{1}{2}.