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 explain why the composite function $\mathrm{fg}$ exists.

Given that $g$ is defined for $x \ge 1$, $g(1) = -0.5$, and $g(x) \to \infty$ as $x$ increases, the function is increasing and its range is $R_g = [-0.5, \infty)$.

The domain of $f$ is given by $x \in \mathbb{R}, x \ne -1$, so $D_f = \mathbb{R} \setminus \{-1\}$.

For the composite function $\mathrm{fg}$ to exist, we require $R_g \subseteq D_f$.
Since $[-0.5, \infty)$ does not contain $-1$, we have $R_g \subseteq D_f$. Therefore, the composite function $\mathrm{fg}$ exists.

2. Find the range of $\mathrm{fg}$ by evaluating the range of $f(x)$ for the domain subset matching $R_g$.

The range of $\mathrm{fg}$ corresponds to the range of $f(x)$ evaluated over $x \in R_g$, which is $x \in [-0.5, \infty)$.

Using a graphing calculator to find the minimum point of $y = e^x + \dfrac{1}{2x+2}$ for $x \ge -0.5$:
The minimum point occurs at $x \approx -0.1634$, yielding a minimum $y$-value of approximately $1.4469$.

As $x \to \infty$, $e^x \to \infty$ and $\dfrac{1}{2x+2} \to 0$, so $f(x) \to \infty$.

3. State the final range to appropriate significant figures.

Therefore, the range of $\mathrm{fg}$ is $[1.4469, \infty)$, which rounds to $[1.45, \infty)$ to 3 significant figures.

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

Using $f(x) = e^x + \frac{1}{2x+2}$, we have:
$fg(x) = e^{g(x)} + \frac{1}{2g(x)+2}$

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

We are given:
$fg(x) = \frac{x}{\sqrt{e}} + \frac{1}{2\ln x + 1}$
Rewrite the first term using $x = e^{\ln x}$ and $\sqrt{e} = e^{0.5}$:
$\frac{x}{\sqrt{e}} = \frac{e^{\ln x}}{e^{0.5}} = e^{\ln x - 0.5}$
Rewrite the denominator of the second term:
$2\ln x + 1 = 2\ln x - 1 + 2 = 2(\ln x - 0.5) + 2$
Substitute these back:
$fg(x) = e^{\ln x - 0.5} + \frac{1}{2(\ln x - 0.5) + 2}$

3. Compare the two expressions to deduce $g(x)$.

Comparing $e^{g(x)} + \frac{1}{2g(x)+2}$ with $e^{\ln x - 0.5} + \frac{1}{2(\ln x - 0.5) + 2}$, we deduce:
$g(x) = \ln x - 0.5$
(Check: $g(1) = \ln 1 - 0.5 = -0.5$, which matches the given condition.)

1. Find the turning point of $f$ to determine the domain where it is one-to-one, and deduce the least integer $k$.

For $f^{-1}$ to exist, $f$ must be a one-to-one function on $x > k$, which means the domain $x > k$ must not contain any turning points.
Using a graphing calculator, the turning point (a local minimum) of $y = e^x + \dfrac{1}{2x+2}$ for $x > -1$ occurs at $x \approx -0.315$.
For $f$ to be strictly increasing, we need $k \ge -0.315$.
The least integer $k$ that satisfies this condition is $0$.

1. Determine the domain and starting point of $f(x)$ based on the restriction from part (c).

From part (c), the least integer $k$ for the inverse to exist is $k=0$, making the domain of $f$ restricted to $x > 0$.
To find the starting point of the graph of $f$, evaluate it at the boundary $x=0$:
$f(0) = e^0 + \frac{1}{2(0)+2} = 1 + 0.5 = 1.5$
Thus, $f(x)$ starts from an open circle at $(0, 1.5)$ and strictly increases. Since $f(x) > x$ for all $x > 0$, the curve lies entirely above the line $y=x$.

2. Determine the domain and starting point of $f^{-1}(x)$ using the properties of inverse functions.

The graph of an inverse function $f^{-1}$ is the reflection of the graph of $f$ across the line $y=x$.
Since the range of $f$ for $x > 0$ is $y > 1.5$, the domain of $f^{-1}$ is $x > 1.5$.
Therefore, the graph of $f^{-1}$ starts from an open circle at $(1.5, 0)$ and strictly increases, lying entirely below the line $y=x$.

3. Identify the equation and domain for the composite function $ff^{-1}$ to complete the requested relationships.

The composite function $ff^{-1}$ is the identity function, mapping an input back to itself, so $ff^{-1}(x) = x$.
However, its domain is restricted to the domain of the inner function $f^{-1}$, which is $x > 1.5$.
Thus, the graph of $ff^{-1}$ is simply the straight line segment $y=x$ restricted to $x > 1.5$, starting from an open circle at $(1.5, 1.5)$.
Drawing this straight line segment directly on the axis of reflection $y=x$ visually demonstrates the symmetric relationship between the curves of $f$ and $f^{-1}$.

1. Set up the inverse relationship and find the corresponding $y$-value on the inverse graph.

Let $y = \mathrm{f}^{-1}(x)$, which implies $x = \mathrm{f}(y)$.
We are given $x = e + \dfrac{1}{4}$. Substituting this into the function gives:
$e + \dfrac{1}{4} = e^y + \dfrac{1}{2y+2}$
By observation, $y = 1$ is a solution. (This is valid as it falls in the restricted domain $x > k$ where $k=0$).

2. Differentiate implicitly with respect to $x$ to find an expression for the gradient $\frac{dy}{dx}$.

Given $x = e^y + \dfrac{1}{2y+2}$, differentiate both sides with respect to $x$:
$1 = \left( e^y - \dfrac{2}{(2y+2)^2} \right) \dfrac{dy}{dx}$
$1 = \left( e^y - \dfrac{1}{2(y+1)^2} \right) \dfrac{dy}{dx}$
$\dfrac{dy}{dx} = \dfrac{1}{e^y - \dfrac{1}{2(y+1)^2}}$

3. Evaluate the gradient at the required point.

Substitute $y = 1$ into the derivative expression to find the gradient of the tangent:
$\dfrac{dy}{dx} = \dfrac{1}{e^1 - \dfrac{1}{2(1+1)^2}}$
$\dfrac{dy}{dx} = \dfrac{1}{e - \dfrac{1}{8}} = \dfrac{8}{8e - 1}$