The function f is defined by $f : x \mapsto \begin{cases} a^2 x - a x^2 & \text{for } x \le \tfrac{1}{2}a, \\ -\tfrac{1}{2}a^2 x + \tfrac{1}{2}a^3 & \text{for } x > \tfrac{1}{2}a. \end{cases}$ where $a$ is a positive constant. The function g is defined by $g : x \mapsto e^x,\; x \le a^3.$

1. Identify the key features of the quadratic portion for $x \le \frac{1}{2}a$.

For $x \le \frac{1}{2}a$, $f(x) = a^2 x - a x^2 = ax(a - x)$.
This forms a downward-facing parabola.
Setting $f(x) = 0$ gives $x = 0$ or $x = a$. Since $x \le \frac{1}{2}a$, the only $x$-intercept in this domain is $(0, 0)$.
The turning point occurs at $x = -\frac{a^2}{2(-a)} = \frac{1}{2}a$.
At $x = \frac{1}{2}a$, $f(\frac{1}{2}a) = a^2(\frac{1}{2}a) - a(\frac{1}{4}a^2) = \frac{1}{4}a^3$.
The curve ends exactly at its maximum turning point $(\frac{1}{2}a, \frac{1}{4}a^3)$.

2. Identify the key features of the linear portion for $x > \frac{1}{2}a$.

For $x > \frac{1}{2}a$, $f(x) = -\frac{1}{2}a^2 x + \frac{1}{2}a^3$.
This is a straight line with a negative gradient.
Setting $f(x) = 0$ gives $\frac{1}{2}a^2 x = \frac{1}{2}a^3 \implies x = a$.
The $x$-intercept is therefore $(a, 0)$, which falls within the domain $x > \frac{1}{2}a$.
Checking continuity at the boundary $x = \frac{1}{2}a$:
$y = -\frac{1}{2}a^2(\frac{1}{2}a) + \frac{1}{2}a^3 = -\frac{1}{4}a^3 + \frac{1}{2}a^3 = \frac{1}{4}a^3$.
The line connects seamlessly to the end of the curve at $(\frac{1}{2}a, \frac{1}{4}a^3)$.

3. Synthesise the features to describe the full sketch of $y = f(x)$.

To sketch the graph on a pair of axes:

• For $x \le \frac{1}{2}a$, draw a concave-down curve originating from the third quadrant, passing through the origin $(0, 0)$, up to the peak at $(\frac{1}{2}a, \frac{1}{4}a^3)$.
• For $x > \frac{1}{2}a$, draw a straight line from the peak downwards, cutting the $x$-axis at $(a, 0)$ and extending into the fourth quadrant.
• Ensure the coordinates $(0,0)$, $(a,0)$, and $(\frac{1}{2}a, \frac{1}{4}a^3)$ are clearly labelled.

1. State the condition required for an inverse function to exist.

For $f^{-1}$ to exist, the function $f$ must be a one-to-one function.

2. Determine the greatest value of $k$ by inspecting the monotonicity of $f(x)$.

From the shape of the graph sketched in part (i), $f(x)$ is strictly increasing for $x \le \frac{1}{2}a$ and strictly decreasing for $x > \frac{1}{2}a$.
To maintain a one-to-one function on a restricted domain of the form $x \le k$, the function must only be strictly increasing and not encompass any part of the decreasing portion.
Therefore, the greatest value of $k$ is $\frac{1}{2}a$.

1. Complete the square to express $x$ in terms of $y$.

From part (ii), $k = \frac{1}{2}a$, so the domain is restricted to $x \le \frac{1}{2}a$.
For $x \le \frac{1}{2}a$, $f(x) = a^2x - ax^2$. Let $y = a^2x - ax^2$.
$y = -a(x^2 - ax)$
$y = -a \left[ \left(x - \frac{1}{2}a\right)^2 - \frac{1}{4}a^2 \right]$
$y = -a\left(x - \frac{1}{2}a\right)^2 + \frac{1}{4}a^3$
$a\left(x - \frac{1}{2}a\right)^2 = \frac{1}{4}a^3 - y$
$\left(x - \frac{1}{2}a\right)^2 = \frac{1}{4}a^2 - \frac{y}{a}$

2. Solve for $x$ and select the correct sign for the square root based on the domain.

Taking the square root of both sides:
$x - \frac{1}{2}a = \pm\sqrt{\frac{1}{4}a^2 - \frac{y}{a}}$
Since the domain is restricted to $x \le \frac{1}{2}a$, we must have $x - \frac{1}{2}a \le 0$. Thus, we reject the positive root and choose the negative square root:
$x - \frac{1}{2}a = -\sqrt{\frac{1}{4}a^2 - \frac{y}{a}}$
$x = \frac{1}{2}a - \sqrt{\frac{1}{4}a^2 - \frac{y}{a}}$

3. Determine the domain of $f^{-1}$.

The domain of $f^{-1}$ is the range of $f$ over the restricted domain $x \le \frac{1}{2}a$.
From the completed square form $y = -a\left(x - \frac{1}{2}a\right)^2 + \frac{1}{4}a^3$, the maximum value is $\frac{1}{4}a^3$ at $x = \frac{1}{2}a$.
Since the parabola opens downwards ($a>0$), the range is $f(x) \le \frac{1}{4}a^3$.
Therefore, the domain of $f^{-1}$ is $x \le \frac{1}{4}a^3$.

4. Express the definition in similar mapping form and state the relationship between the graphs.

Replacing $y$ with $x$ to define the inverse function mapping:
$f^{-1} : x \mapsto \frac{1}{2}a - \sqrt{\frac{1}{4}a^2 - \frac{x}{a}} \quad \text{for } x \le \frac{1}{4}a^3$

Relationship:
The graphs of $y = f(x)$ and $y = f^{-1}(x)$ are reflections of each other in the line $y = x$.

1. Determine the range of f, compare it with the domain of g, and state the conclusion.

From the definition of $f$, the maximum value occurs at $x = \frac{1}{2}a$ (which would also be visible from the graph sketched in part (i)).
$f\left(\frac{1}{2}a\right) = a^2\left(\frac{1}{2}a\right) - a\left(\frac{1}{2}a\right)^2 = \frac{1}{4}a^3$
Thus, the range of $f$ is $R_f = (-\infty, \frac{1}{4}a^3]$.
The domain of $g$ is given as $D_g = (-\infty, a^3]$.
Since $a$ is a positive constant, $a^3 > 0$, which implies that $\frac{1}{4}a^3 < a^3$.
Therefore, $R_f \subseteq D_g$.
Since the range of $f$ is a subset of the domain of $g$, the composite function $gf$ exists.

1. State the range of the inner function $f$.

From the definition or graph of $f(x)$, the maximum point occurs at $x = \frac{1}{2}a$ giving $f(\frac{1}{2}a) = \frac{1}{4}a^3$.
As $x \to \pm\infty$, $f(x) \to -\infty$.
Thus, the range of $f$ is $R_f = (-\infty, \frac{1}{4}a^3]$.

2. Apply the outer function $g$ to the range of $f$ to find the range of $gf$.

The range of the composite function $gf$ is found by substituting $R_f$ into $g(x) = e^x$.
Since $g$ is a strictly increasing function:
The maximum value is $g(\frac{1}{4}a^3) = e^{\frac{1}{4}a^3}$.
As the input tends to $-\infty$, $e^x \to 0$.
Therefore, the range of $gf$ is $(0, e^{\frac{1}{4}a^3}]$.