The functions $f$ and $g$ are defined by $$f: x ightarrow 2|x| + 3, \, x \in \mathbb{R},$$ $$g: x \rightarrow 3 - 4x, \, x \in \mathbb{R}.$$ (a) State the range of $f$ - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 8

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

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

Then we calculate $f(g(1)) = f(-1)$:

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

So, $fg(1) = 5$.

To find the inverse function $g^{-1}$, we start from the equation of $g(x)$:

$y = 3 - 4x.$

To find $g^{-1}(y)$, we rearrange this equation to express $x$ in terms of $y$:

1. Rearranging gives:
   $4x = 3 - y$
2. Dividing by 4 results in:
   $x = \frac{3 - y}{4}.$

Thus, the inverse function is:

$g^{-1}(y) = \frac{3 - y}{4}$

In terms of $x$, this can be expressed as:

$g^{-1}(x) = \frac{3 - x}{4}.$

Start by defining $g(x)$ as $g(x) = 3 - 4x$. We can substitute into the original equation:

1. Calculate $g(g(x))$:
   $g(g(x)) = g(3 - 4x) = 3 - 4(3 - 4x) = 3 - 12 + 16x = 16x - 9.$

2. Substitute into the equation:
   $gg(x) + [g(x)]^2 = (16x - 9) + (3 - 4x)^2 = 0.$

3. Expand $[g(x)]^2$:
   $[g(x)]^2 = (3 - 4x)^2 = 9 - 24x + 16x^2.$

4. Combine terms: $gg(x) + [g(x)]^2 = (16x - 9) + 9 - 24x + 16x^2 = 16x^2 - 8x = 0.$

5. Factor the quadratic: $8x(2x - 1) = 0.$

6. Set each factor to zero:

• $8x = 0$ leads to $x = 0$.
• $2x - 1 = 0$ leads to $x = \frac{1}{2}.$

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