已知 $y = x(2x-4)^3$。

1. Differentiate the function $y$ with respect to $x$ using the product and chain rules.

$\frac{dy}{dx} = 1 \cdot (2x-4)^3 + x \cdot 3(2x-4)^2 \cdot 2$

2. Simplify and factorise the derivative.

$\frac{dy}{dx} = (2x-4)^3 + 6x(2x-4)^2$
$\frac{dy}{dx} = (2x-4)^2 [ (2x-4) + 6x ]$
$\frac{dy}{dx} = (2x-4)^2 (8x - 4)$

3. Set the derivative to zero to find the $x$-coordinates of the stationary points.

For stationary points, $\frac{dy}{dx} = 0$:
$(2x-4)^2 (8x - 4) = 0$
$2x - 4 = 0 \implies x = 2$
$8x - 4 = 0 \implies x = 0.5$

4. Find the corresponding $y$-coordinates for each $x$-value.

When $x = 2$:
$y = 2(2(2)-4)^3 = 2(0)^3 = 0$

When $x = 0.5$:
$y = 0.5(2(0.5)-4)^3 = 0.5(1-4)^3 = 0.5(-3)^3 = 0.5(-27) = -13.5$

1. Recall the first derivative and the $x$-coordinates of the stationary points from part (i)

From part (i),
$\frac{dy}{dx} = (2x-4)^2(8x-4)$
The stationary points are at $x = 0.5$ and $x = 2$.

2. Use the first derivative test to determine the nature of the stationary point at $x = 0.5$

For $x = 0.5$:
When $x = 0.4$, $\frac{dy}{dx} = (2(0.4)-4)^2(8(0.4)-4) = (-3.2)^2(-0.8) < 0$
When $x = 0.5$, $\frac{dy}{dx} = 0$
When $x = 0.6$, $\frac{dy}{dx} = (2(0.6)-4)^2(8(0.6)-4) = (-2.8)^2(0.8) > 0$
Since the gradient changes from negative to positive, the point at $x = 0.5$ is a minimum point.

3. Use the first derivative test to determine the nature of the stationary point at $x = 2$

For $x = 2$:
When $x = 1.9$, $\frac{dy}{dx} = (2(1.9)-4)^2(8(1.9)-4) = (-0.2)^2(11.2) > 0$
When $x = 2$, $\frac{dy}{dx} = 0$
When $x = 2.1$, $\frac{dy}{dx} = (2(2.1)-4)^2(8(2.1)-4) = (0.2)^2(12.8) > 0$
Since the gradient is positive on both sides, the point at $x = 2$ is a stationary point of inflexion.