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 5 - 2013 - Paper 8

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

$g(1) = 3 - 4(1) = 3 - 4 = -1$

Next, we substitute $g(1)$ into $f$:

$f(g(1)) = f(-1) = 2|-1| + 3 = 2(1) + 3 = 2 + 3 = 5$

Therefore, $fg(1) = 5$.

To find the inverse of the function $g(x) = 3 - 4x$, we start by replacing $g(x)$ with $y$:

$y = 3 - 4x$

Now, solving for $x$, we rearrange:

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

Switching $x$ and $y$ for the inverse gives:

g^{-1}(x) = rac{3 - x}{4}

First, we compute $g(g(x))$. Start by finding $g(x)$:

$g(x) = 3 - 4x$

Now substitute $g(x)$ into itself:

$g(g(x)) = g(3 - 4x) = 3 - 4(3 - 4x) = 3 - 12 + 16x = 16x - 9$

Now substituting into the equation gives:

$gg(x) + [g(x)]^2 = 0$ $16x - 9 + (3 - 4x)^2 = 0$

Expanding $(3 - 4x)^2$:

$(3 - 4x)^2 = 9 - 24x + 16x^2$

Putting it back into the equation:

$16x - 9 + 9 - 24x + 16x^2 = 0$ $16x^2 - 8x = 0$ $8x(2x - 1) = 0$

Thus, the solutions are: x = 0 ext{ or } x = rac{1}{2}