The function $f$ is defined by $f : x \mapsto x^2 - 4x,\ x \in \mathbb{R},\ |x|<1.$ The function $g$ is defined by $g : x \mapsto 1 + \frac{1}{x+3},\ x \in \mathbb{R},\ x \ne -3.$ The function $h$ is defined by $h : x \mapsto gf(x),\ x \in \mathbb{R},\ |x|<1.$

1. Show that the function is one-to-one by completing the square and analysing its shape over the given domain.

$f(x) = x^2 - 4x = (x - 2)^2 - 4$
The graph of $y = f(x)$ is a parabola opening upwards with its line of symmetry at $x = 2$.
Since the domain is $|x| < 1$, or $-1 < x < 1$, this region lies entirely to the left of the turning point. Thus, $f$ is strictly decreasing on this domain.
Because $f$ is strictly decreasing, any horizontal line $y = k$ intersects the curve at most once. Therefore, $f$ is a one-to-one function, which means $f^{-1}$ exists.

2. Set up an equation to find the rule for the inverse function.

Let $y = f(x)$.
$y = (x - 2)^2 - 4$
$(x - 2)^2 = y + 4$
$x - 2 = \pm\sqrt{y + 4}$
$x = 2 \pm\sqrt{y + 4}$

3. Select the correct algebraic sign based on the domain restriction.

Since the domain of $f$ is $-1 < x < 1$, we must have $x < 2$.
This requires us to reject the positive square root branch.
$x = 2 - \sqrt{y + 4}$

4. Determine the domain of the inverse function and express it in mapping notation.

The domain of $f^{-1}$ is exactly the range of $f$.
Because $f$ is strictly decreasing on $-1 < x < 1$, its range is bounded by the values at the endpoints:
$f(1) < f(x) < f(-1)$
$1^2 - 4(1) < f(x) < (-1)^2 - 4(-1)$
$-3 < f(x) < 5$
So, the domain of $f^{-1}$ is $-3 < x < 5$.
Writing this in a similar form to the question:
$f^{-1}: x \mapsto 2 - \sqrt{x+4}, \ x \in \mathbb{R}, \ -3 < x < 5$

1. Find the range of the function $f$.

Let $f(x) = x^2 - 4x = (x-2)^2 - 4$.
For the domain $-1 < x < 1$, the function is on the left of its minimum point at $x=2$, so it is strictly decreasing.
Evaluating the boundaries:
$f(-1) = (-1)^2 - 4(-1) = 5$
$f(1) = 1^2 - 4(1) = -3$
Therefore, the range of $f$ is $R_f = (-3, 5)$.

2. State the domain of $g$ and check the condition for the existence of $gf$.

The domain of $g$ is given by $x \neq -3$, so $D_g = \mathbb{R} \setminus \{-3\}$.
For the composite function $gf$ to exist, we require $R_f \subseteq D_g$.
Since all values in $R_f = (-3, 5)$ are strictly greater than $-3$, it follows that $-3 \notin R_f$.
Thus, $R_f \subseteq D_g$, and the composite function $gf$ exists.

1. Substitute $f(x)$ into $g(x)$ to find the composite function $gf(x)$.

$gf(x) = g(x^2 - 4x)$
$= 1 + \frac{1}{(x^2 - 4x) + 3}$
$= 1 + \frac{1}{x^2 - 4x + 3}$

1. Find the first derivative of $h(x)$.

From part (iii), $h(x) = gf(x) = 1 + (x^2 - 4x + 3)^{-1}$.
Differentiating with respect to $x$:
$h'(x) = -1(x^2 - 4x + 3)^{-2}(2x - 4)$
$h'(x) = \frac{4 - 2x}{(x^2 - 4x + 3)^2}$

2. Analyse the sign of $h'(x)$ for the given domain.

The domain of $h$ is $|x| < 1$, which means $-1 < x < 1$.
For all $x$ in this domain:

• The denominator $(x^2 - 4x + 3)^2 > 0$ (since $x \neq 1$ and $x \neq 3$).
• For the numerator, since $x < 1$, we have $2x < 2 \implies 4 - 2x > 2 > 0$.
  Since both the numerator and denominator are positive, $h'(x) > 0$ for all $x \in (-1, 1)$.

3. Conclude the behaviour of $h(x)$.

Since $h'(x) > 0$ for all $x$ in its domain, $h$ is strictly increasing.

4. Evaluate the limits of $h(x)$ at the domain endpoints to find the range.

Since $h$ is strictly increasing, its range is bounded by its limits as $x \to -1^+$ and $x \to 1^-$.
As $x \to -1^+$,
$h(x) \to 1 + \frac{1}{(-1)^2 - 4(-1) + 3} = 1 + \frac{1}{8} = \frac{9}{8}$
As $x \to 1^-$, the quadratic can be factored as $(x^2 - 4x + 3) = (x-1)(x-3)$.
Substituting $x \to 1^-$, we see $(x-1) \to 0^-$ and $(x-3) \to -2$, so their product $(x-1)(x-3) \to 0^+$.
Thus, $\frac{1}{x^2 - 4x + 3} \to \infty$, which means $h(x) \to \infty$.

5. State the final range of $h$.

Therefore, the range of $h$ is $\left(\frac{9}{8}, \infty\right)$.