The function f is defined by $f : x \mapsto e^{x} + \dfrac{1}{2x+2}\quad\text{for } x \in \mathbb{R},\ x \ne -1.$ A function g, defined for $x \in \mathbb{R},\ x \ge 1$, is such that $y \to \infty$ as $x$ increases. It is also given that $g(1) = -0.5$.

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

Given $g(1) = -0.5$ and $y \to \infty$ as $x$ increases, the range of $g$ is $R_g = [-0.5, \infty)$.
From the definition of $f$, its domain is $D_f = \mathbb{R} \setminus \{-1\}$.

2. Explain why $fg$ exists based on $R_g$ and $D_f$.

For the composite function $fg$ to exist, we require $R_g \subseteq D_f$.
Since $-1 \notin [-0.5, \infty)$, the condition $R_g \subseteq D_f$ is strictly satisfied, hence $fg$ exists.

3. Find the range of $fg$ by analyzing the values of $f(x)$ over the domain $R_g$.

The range of $fg$ corresponds to the range of $f(x)$ for the restricted domain $x \in R_g$, i.e., $x \ge -0.5$.
Using a graphing calculator to find the minimum point of $y = e^x + \dfrac{1}{2x+2}$ for $x \ge -0.5$:
The minimum occurs at $x \approx -0.214$, with a minimum value of $y \approx 1.4435$.
As $x \to \infty$, $f(x) \to \infty$.
Therefore, the range of $fg$ is $[1.44, \infty)$ (correct to 3 s.f.).

1. Express $fg(x)$ using the definition of $f(x)$.

$fg(x) = f(g(x)) = e^{g(x)} + \frac{1}{2g(x)+2}$

2. Rewrite the given expression for $fg(x)$ to match the form of $f(x)$.

Given $fg(x) = \frac{x}{\sqrt{e}} + \frac{1}{2\ln x + 1}$
Rewrite the first term using indices and logarithms:
$\frac{x}{\sqrt{e}} = x e^{-0.5} = e^{\ln x} e^{-0.5} = e^{\ln x - 0.5}$
Rewrite the denominator of the second term to reveal the factor of $2$:
$2\ln x + 1 = 2\ln x - 1 + 2 = 2(\ln x - 0.5) + 2$

Substituting these back gives:
$fg(x) = e^{\ln x - 0.5} + \frac{1}{2(\ln x - 0.5) + 2}$

3. Deduce $g(x)$ by comparing both expressions.

Comparing $e^{g(x)} + \frac{1}{2g(x)+2}$ with $e^{\ln x - 0.5} + \frac{1}{2(\ln x - 0.5) + 2}$, we can deduce by observation that:
$g(x) = \ln x - 0.5$
(Check: $g(1) = \ln 1 - 0.5 = -0.5$, which satisfies the given condition $g(1) = -0.5$. Furthermore, $y \to \infty$ as $x$ increases, satisfying the other given property.)

1. Set up the condition for the inverse function to exist and find the derivative of $f(x)$.

For $f^{-1}$ to exist, $f$ must be a one-to-one function on its domain $x > k$. This implies its derivative $f'(x)$ should not change sign.
$f'(x) = e^x - \frac{1}{2(x+1)^2}$

2. Analyse the sign of $f'(x)$ to determine if $k=0$ and $k=-1$ are valid domains.

Let us check if $k = 0$ works. For $x > 0$:
$e^x > e^0 = 1 \quad \text{and} \quad \frac{1}{2(x+1)^2} < \frac{1}{2(0+1)^2} = 0.5$
Therefore, $f'(x) > 1 - 0.5 = 0.5 > 0$. Thus, $f(x)$ is strictly increasing and one-to-one for $x > 0$.

Let us check if $k = -1$ works. For $x > -1$, we can evaluate $f'(x)$ at $x = -0.5$:
$f'(-0.5) = e^{-0.5} - \frac{1}{2(0.5)^2} \approx -1.39 < 0$
Since $f'(0) = 0.5 > 0$ and $f'(-0.5) < 0$, $f'(x)$ changes sign, meaning $f(x)$ has a turning point and is not one-to-one for $x > -1$.

(Note: Using a graphing calculator confirms a minimum turning point at $x \approx -0.214$, which means we must restrict the domain to $x \ge -0.214$.)

3. State the least value of the integer $k$.

Since $k$ must be an integer and $k \ge -0.214$, the least integer value is $0$.

1. Determine the endpoints and shapes for the graph of $f$.

From part (c), the domain of $f$ is restricted to $x > 0$ (since $k=0$).
Evaluating $f(x)$ at the boundary $x = 0$:
$f(0) = e^0 + \frac{1}{2(0)+2} = 1.5$
Thus, the range of $f$ is $y > 1.5$.
The graph of $y = f(x)$ is a strictly increasing curve starting from an empty circle at $(0, 1.5)$.

2. Determine the features of $f^{-1}$ using reflection properties.

The graph of $y = f^{-1}(x)$ is the exact reflection of the graph of $y = f(x)$ in the line $y = x$.
Therefore, its domain is $x > 1.5$ and its range is $y > 0$.
It starts from an empty circle at $(1.5, 0)$ and is strictly increasing (concave downwards).

3. Identify the equation and domain for the graph of $ff^{-1}$ and state the graphical relationship.

By definition, the composite function $ff^{-1}(x) = x$.
Its domain must match the domain of the inner function $f^{-1}$, which is the range of $f$ ($x > 1.5$).
So, the graph of $y = ff^{-1}(x)$ is the straight line $y = x$ restricted to $x > 1.5$, starting from an empty circle at $(1.5, 1.5)$.
To show the relationship clearly, the line $y = x$ is drawn dashed for $x \le 1.5$ as the axis of symmetry between $f$ and $f^{-1}$.

1. Find the corresponding coordinate on the curve $y = f(x)$ that maps to $e + \dfrac{1}{4}$.

Let $y = f^{-1}\left(e + \dfrac{1}{4}\right)$. This implies that $f(y) = e + \dfrac{1}{4}$.
$e^y + \dfrac{1}{2y+2} = e + \dfrac{1}{4}$
By observation, we can see that $y = 1$ is a solution. (This is unique as $f$ is one-to-one on the restricted domain).

2. Differentiate $f(x)$ with respect to $x$.

Express $f(x)$ with negative indices for easier differentiation:
$f(x) = e^x + \dfrac{1}{2}(x+1)^{-1}$
$f'(x) = e^x - \dfrac{1}{2}(x+1)^{-2} = e^x - \dfrac{1}{2(x+1)^2}$

3. Evaluate the gradient of the tangent to $f^{-1}$ at the given point using the inverse derivative rule.

For $y = f^{-1}(x)$, we have $x = f(y)$. Differentiating implicitly with respect to $x$ gives:
$1 = f'(y) \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{1}{f'(y)}$
When $x = e + \dfrac{1}{4}$, $y = 1$.
Substitute $y = 1$ into $f'$:
$f'(1) = e^1 - \dfrac{1}{2(1+1)^2} = e - \dfrac{1}{8}$
The gradient of the tangent is therefore:
$\dfrac{dy}{dx} = \dfrac{1}{e - \frac{1}{8}} = \dfrac{8}{8e - 1}$