The functions $f$ and $g$ are defined by $f : x \mapsto \sqrt{16 - 4x},\; x \in \mathbb{R},\; x \le 4,$ $g : x \mapsto x^{2},\; x \in \mathbb{R}.$

1. Find the point of intersection that lies on the line of symmetry $y = x$.

$f(x) = x \implies \sqrt{16 - 4x} = x$
Squaring both sides:
$16 - 4x = x^2 \implies x^2 + 4x - 16 = 0$
Using the quadratic formula:
$x = \frac{-4 \pm \sqrt{16 - 4(1)(-16)}}{2} = \frac{-4 \pm \sqrt{80}}{2} = -2 \pm 2\sqrt{5}$
Since the principal square root is non-negative, we must have $x \ge 0$. Thus, we reject $-2 - 2\sqrt{5}$.
$x = -2 + 2\sqrt{5}$
One point of intersection is $(-2 + 2\sqrt{5}, -2 + 2\sqrt{5})$.

2. Find any additional points of intersection by considering the intercepts of $f(x)$.

For the $y$-intercept of $f$, set $x = 0$:
$f(0) = \sqrt{16 - 0} = 4 \implies (0, 4) \text{ is on } f$
For the $x$-intercept of $f$, set $y = 0$:
$\sqrt{16 - 4x} = 0 \implies x = 4 \implies (4, 0) \text{ is on } f$
Since the graph of $f^{-1}$ is the reflection of $f$ across the line $y = x$, any point $(a, b)$ on $f$ means $(b, a)$ is on $f^{-1}$.
Since $(0, 4)$ is on $f$, its reflection $(4, 0)$ is on $f^{-1}$.
Since $(4, 0)$ is on $f$, its reflection $(0, 4)$ is on $f^{-1}$.
Therefore, the symmetric points $(0, 4)$ and $(4, 0)$ lie on both graphs and are valid intersection points.

3. Sketch the graphs of $f$ and $f^{-1}$ on the same diagram.

The domain of $f$ is $x \le 4$ and its range is $y \ge 0$.
Therefore, the domain of $f^{-1}$ is $x \ge 0$ and its range is $y \le 4$.

• Draw $f$ starting from $(4, 0)$, curving upwards to the left, passing through $(0, 4)$.
• Draw $f^{-1}$ starting from $(0, 4)$, curving downwards to the right, passing through $(4, 0)$.
• Ensure the curves smoothly intersect at $(0, 4)$, $(4, 0)$, and $(-2+2\sqrt{5}, -2+2\sqrt{5})$.

(A standard graph sketch is drawn matching these features)

1. State the range of $g$ and the domain of $f$.

Range of $g$, $R_g = [0, \infty)$
Domain of $f$, $D_f = (-\infty, 4]$

2. Explain why $fg$ does not exist by checking the subset condition.

For the composite function $fg$ to exist, the range of $g$ must be a subset of the domain of $f$ ($R_g \subseteq D_f$).
Since $[0, \infty) \not\subseteq (-\infty, 4]$ (for example, $5 \in R_g$ but $5 \notin D_f$), the condition is not satisfied.
Hence, $fg$ does not exist.

1. Substitute $f(x)$ into $g(x)$ to find the algebraic rule for $gf(x)$.

$gf(x) = g(f(x))$
$gf(x) = g(\sqrt{16 - 4x})$
$gf(x) = (\sqrt{16 - 4x})^2$
$gf(x) = 16 - 4x$

2. Express the composite function $gf$ in the requested mapping form, including its domain.

The domain of a composite function $gf$ is the set of all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$.
Since the domain of $g$ is $x \in \mathbb{R}$, the domain of $gf$ is simply the domain of $f$, which is $x \le 4$.

In similar form:
$gf : x \mapsto 16 - 4x, \; x \in \mathbb{R}, \; x \le 4$

3. Determine the range of the composite function $gf$.

To find the range, we observe the output of $gf(x)$ for its domain $x \le 4$.
$x \le 4$
$4x \le 16$
$16 - 4x \ge 0$
Therefore, the range of $gf$ is $gf(x) \ge 0$.

(Alternatively, the range of $f(x)$ is $[0, \infty)$. Substituting $y \in [0, \infty)$ into $g(y) = y^2$ yields $[0, \infty)$).