The function f, with domain the set of all real values, is given by $\text{f}(x) = \begin{cases} 4 - x^2 & \text{for } 0 < x \le 2, \\ 2x - 4 & \text{for } 2 < x \le 4, \end{cases}$ and that $\text{f}(x) = \text{f}(x+4)$.

1. Use the periodic property of the function to reduce the argument to within the given base interval.

Since $\text{f}(x) = \text{f}(x+4)$ for all real $x$, the function is periodic with a period of 4.
$\text{f}(45) = \text{f}(45 - 11 \times 4) = \text{f}(1)$

2. Evaluate the function using the appropriate piece of the definition.

For $x = 1$, it falls in the interval $0 < x \le 2$, so we use the definition $\text{f}(x) = 4 - x^2$.
$\text{f}(1) = 4 - (1)^2$
$\text{f}(1) = 3$

1. Analyse the shape and key coordinates of the function in its base interval $(0, 4]$.

For the interval $0 < x \le 4$:

• $y = 4 - x^2$ for $0 < x \le 2$: This is a concave down parabola joining $(0, 4)$ to $(2, 0)$.
• $y = 2x - 4$ for $2 < x \le 4$: This is a straight line starting from $(2, 0)$ and ending at $(4, 4)$.

The segments connect seamlessly at $(2,0)$.

2. Apply the periodic property to determine the key points and shapes for the full sketching domain $[-4, 7]$.

The property $\text{f}(x) = \text{f}(x+4)$ means the function is periodic with a period of $4$. We translate the base pattern left and right to cover $-4 \le x \le 7$:

For $-4 \le x \le 0$ (shifted left by $4$):

• Parabola from $(-4, 4)$ to $(-2, 0)$
• Straight line from $(-2, 0)$ to $(0, 4)$

For $4 \le x \le 7$ (shifted right by $4$):

• Parabola from $(4, 4)$ to $(6, 0)$
• Straight line from $(6, 0)$ to $(7, 2)$ (since $\text{f}(7) = \text{f}(3) = 2(3) - 4 = 2$)

3. State the range of the function.

Observing the behaviour across one complete period, the minimum value attained is $0$ (at $x = -2, 2, 6$) and the maximum value is $4$ (at $x = -4, 0, 4$).

Range of f is $0 \le \text{f}(x) \le 4$.

1. State the range of f and domain of g

From part (ii), the range of $\text{f}$ is $R_{\text{f}} = [0, 4]$.
From the given definition of $\text{g}$, the domain of $\text{g}$ is $D_{\text{g}} = (0, 1)$.

2. Explain why the composite function does not exist

For the composite function $\text{gf}$ to exist, we require $R_{\text{f}} \subseteq D_{\text{g}}$.
Since $[0, 4] \not\subseteq (0, 1)$, the condition is not satisfied and therefore $\text{gf}$ does not exist.

1. Find the range of g(x) for its given domain to identify which rule of f applies.

For $0 < x < 1$, we have $-1 < x-1 < 0$.
Squaring gives $0 < (x-1)^2 < 1$.
Adding $2$ gives $2 < (x-1)^2 + 2 < 3$.
So, the range of $\text{g}$ is $2 < \text{g}(x) < 3$.
Since this lies within the interval $2 < x \le 4$, we must use the rule $\text{f}(x) = 2x - 4$.

2. Substitute g(x) into the appropriate rule of f to find an expression for fg(x).

$\text{fg}(x) = \text{f}(\text{g}(x)) = 2(\text{g}(x)) - 4$
$\text{fg}(x) = 2((x-1)^2 + 2) - 4$
$\text{fg}(x) = 2(x-1)^2 + 4 - 4$
$\text{fg}(x) = 2(x-1)^2$

3. Equate the expression for fg(x) to 1/2 to find the required value.

Let $x = (\text{fg})^{-1}\left(\frac{1}{2}\right)$. Then $\text{fg}(x) = \frac{1}{2}$.
$2(x-1)^2 = \frac{1}{2}$
$(x-1)^2 = \frac{1}{4}$
$x - 1 = \pm \frac{1}{2}$

4. Solve for x and reject any value outside the domain of g.

$x = 1 + \frac{1}{2} = \frac{3}{2} \quad \text{or} \quad x = 1 - \frac{1}{2} = \frac{1}{2}$
Since the domain of $\text{g}$ (and thus $\text{fg}$) is $0 < x < 1$, we reject $x = \frac{3}{2}$.
$(\text{fg})^{-1}\left(\frac{1}{2}\right) = \frac{1}{2}$