The functions f and g are defined by $f:x\mapsto |(x+1)(x-3)|,\quad x\in\mathbb{R},$ $g:x\mapsto x-2,\quad x\in\mathbb{R}\ \text{and } x\ge -1.$

1. Determine the range of $f$.

The function $f$ is given by $f(x) = |(x+1)(x-3)|$. Since the absolute value of any real expression is non-negative, and $f(-1) = 0$, the range of $f$ is the set of all non-negative real numbers.
$R_f = [0, \infty)$

2. Compare the range of $f$ with the domain of $g$ to prove existence.

The domain of $g$ is given as $x \ge -1$, so $D_g = [-1, \infty)$.
For the composite function $gf$ to exist, the range of $f$ must be a subset of the domain of $g$.
Since $R_f \subseteq D_g$ (i.e., $[0, \infty) \subseteq [-1, \infty)$), the composite function $gf$ exists.

3. Determine the range of $gf$.

The range of $gf$ is the image of $R_f$ under the function $g$.
We evaluate $g(x) = x - 2$ for $x \in R_f$, which means for $x \ge 0$.
If $x \ge 0$, then $x - 2 \ge -2$.
Therefore, the range of $gf$ is $[-2, \infty)$.

1. Show that $f$ is not a one-one function by providing a counterexample.

Substitute $x = -1$ and $x = 3$ into $f(x)$:
$f(-1) = |(-1+1)(-1-3)| = 0$
$f(3) = |(3+1)(3-3)| = 0$

Since $f(-1) = f(3)$ but $-1 \neq 3$, multiple inputs map to the same output. Therefore, $f$ is not a one-one function.

2. Determine the largest value of $k$ for which $f$ is one-one for $x \le k$.

The curve $y = (x+1)(x-3)$ is a parabola opening upwards with $x$-intercepts at $x = -1$ and $x = 3$.
Taking the modulus reflects the region below the $x$-axis (between $x = -1$ and $x = 3$) above the $x$-axis.
From the shape of the graph of $f(x)$, the function is strictly decreasing for $x \le -1$ and strictly increasing for $-1 \le x \le 1$.
For $f(x)$ to remain one-one on the restricted domain $x \le k$, the interval cannot extend past the turning point at $x = -1$.
Hence, the largest value of $k$ is $-1$.

1. State the expression for $f(x)$ on the restricted domain $x \le -1$ without the modulus.

From part (ii), the restricted domain for $f$ to be one-one is $x \le -1$.
For $x \le -1$, we have $(x+1) \le 0$ and $(x-3) \le -4 < 0$, so their product is non-negative: $(x+1)(x-3) \ge 0$.
Thus, the modulus can be removed directly:
$f(x) = (x+1)(x-3) = x^2 - 2x - 3$

2. Make $x$ the subject of the formula, selecting the correct root based on the domain.

Let $y = x^2 - 2x - 3$.
Completing the square gives:
$y = (x-1)^2 - 4$
$(x-1)^2 = y + 4$
$x - 1 = \pm\sqrt{y+4}$
Since the domain is $x \le -1$, it follows that $x - 1 \le -2$. Therefore, we must choose the negative square root to satisfy this condition:
$x - 1 = -\sqrt{y+4}$
$x = 1 - \sqrt{y+4}$

3. Determine the domain of $f^{-1}$ and write the final definition in mapping notation.

The domain of $f^{-1}$ is exactly the range of $f$ for $x \le -1$.
At $x = -1$, $f(-1) = 0$. Since $f$ is strictly decreasing on the interval $(-\infty, -1]$, its values range from $0$ up to $\infty$. Thus, the range of $f$ is $y \ge 0$.
Hence, the domain of $f^{-1}$ is $x \ge 0$.
Defining $f^{-1}$ in a similar form to the original functions:
$f^{-1}: x \mapsto 1 - \sqrt{x+4}, \quad x \in \mathbb{R}, x \ge 0$