The function f is defined by $f : x \mapsto \dfrac{5}{(x-4)^2}$ for $x\in\mathbb{R},\ x<4$.

1. Set $y = f(x)$ and rearrange to make $(x-4)^2$ the subject.

Let $y = \frac{5}{(x-4)^2}$.
$(x-4)^2 = \frac{5}{y}$

2. Take the square root, choose the correct sign based on the domain of $f$, and express $f^{-1}(x)$.

$x - 4 = \pm\sqrt{\frac{5}{y}}$
Since the domain of $f$ is $x < 4$, it follows that $x - 4 < 0$. Therefore, we must reject the positive root.
$x - 4 = -\sqrt{\frac{5}{y}}$
$x = 4 - \sqrt{\frac{5}{y}}$
Hence, $f^{-1}(x) = 4 - \sqrt{\frac{5}{x}}$.

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

The domain of $f^{-1}$ is equal to the range of $f$.
For $x < 4$, $(x-4)^2 > 0$.
Therefore, $f(x) = \frac{5}{(x-4)^2} > 0$.
The range of $f$ is $(0, \infty)$.
Thus, the domain of $f^{-1}$ is $x > 0$ (or $x \in \mathbb{R}, x > 0$).

1. Identify the asymptotes and axis intercepts for the graph of $y = f(x)$.

Given $f(x) = \dfrac{5}{(x-4)^2}$ for $x < 4$:

• As $x \to 4^-$, $y \to \infty$, so the vertical asymptote is $x = 4$.
• As $x \to -\infty$, $y \to 0$, so the horizontal asymptote is $y = 0$.
• When $x = 0$, $y = \dfrac{5}{(0-4)^2} = \dfrac{5}{16}$, giving a $y$-intercept of $(0, \dfrac{5}{16})$.

2. Determine the corresponding features for the graph of $y = f^{-1}(x)$ by reflection.

The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$. We map the features by swapping the $x$ and $y$ coordinates:

• The vertical asymptote $x = 4$ becomes the horizontal asymptote $y = 4$.
• The horizontal asymptote $y = 0$ becomes the vertical asymptote $x = 0$ (the positive $y$-axis).
• The $y$-intercept $(0, \dfrac{5}{16})$ becomes the $x$-intercept $(\dfrac{5}{16}, 0)$.

3. Sketch the graphs on a single diagram, showing the reflection and intersections.

Draw the axes and the line $y = x$.
Plot the curve $y = f(x)$ passing through $(0, \dfrac{5}{16})$ and approaching the lines $x=4$ and $y=0$.
Plot the curve $y = f^{-1}(x)$ passing through $(\dfrac{5}{16}, 0)$ and approaching the lines $y=4$ and $x=0$.
Ensure the curves intersect each other exclusively on the line $y = x$ (as they will intersect at two points, though calculating exact coordinates is left for part iii).
Label all curves, asymptotes, and intercepts clearly.

1. Set up the equation for the intersection points.

At the points of intersection of $y=\mathrm{f}(x)$ and $y=\mathrm{f}^{-1}(x)$, the curves intersect on the line $y=x$.
$\mathrm{f}(x) = x$
$\frac{5}{(x-4)^2} = x$

2. Rearrange the equation to show the given cubic equation.

$5 = x(x-4)^2$
$5 = x(x^2 - 8x + 16)$
$x^3 - 8x^2 + 16x - 5 = 0 \quad \text{(shown)}$

3. Factorise the cubic equation to find its roots.

Let $P(x) = x^3 - 8x^2 + 16x - 5$.
By trial and error (or using a calculator), we can find an integer root:
$P(5) = 125 - 200 + 80 - 5 = 0$
So, $(x-5)$ is a factor.
By long division or comparing coefficients:
$x^3 - 8x^2 + 16x - 5 = (x-5)(x^2 - 3x + 1) = 0$

4. Solve for the exact values of $x$, applying the domain restriction.

$(x-5)(x^2 - 3x + 1) = 0$
$x = 5 \text{ or } x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2}$
$x = 5 \text{ or } x = \frac{3 \pm \sqrt{5}}{2}$
Since the domain of $\mathrm{f}(x)$ is restricted to $x < 4$, we reject $x = 5$.
The exact $x$-coordinates are $\frac{3 - \sqrt{5}}{2}$ and $\frac{3 + \sqrt{5}}{2}$.