The stationary points of a curve are the points where the first derivative is zero.
The nature of the stationary point can be determined by the second derivative. If the second derivative is positive, the stationary point is a local minimum. If the second derivative is negative, the stationary point is a local maximum.
Solution: The given curve $C$ is defined by the equation
\begin{equation}\label{eq:01Mar2025-1}
y = x^2 - 2x - 24 \sqrt{x}.
\end{equation}
Differentiating equation \eqref{eq:01Mar2025-1} with respect to $x$, we get
\begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x} & = 2x - 2 - 12 \left( x \right)^{-\frac{1}{2}} \\
& = 2x - 2 - \frac{12}{\sqrt{x}}.
\end{align*}
For the second derivative, $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$, differentiate the expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$ with respect to $x$ to get
\begin{align*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = 2 + 6 x^{-\frac{3}{2}} = 2 + \frac{6}{x^{3/2}}.
\end{align*}
To verify that $C$ has a stationary point when $x = 4$, we need to check if $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ at $x = 4$. Substituting $x = 4$ into $\frac{\mathrm{d}y}{\mathrm{d}x}$, we get
\begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{x=4} & = 2(4) - 2 - \frac{12}{\sqrt{4}} \\
& = 8 - 2 - 6 \\
& = 0.
\end{align*}
This confirms that there is a stationary point at $x = 4$.
Next, we check the second derivative at the stationary point to determine its nature:
\begin{align*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg|_{x=4} & = 2 + 6(4)^{-\frac{3}{2}} \\
& = 2 + 6 \cdot \frac{1}{8} = \frac{11}{4} > 0.
\end{align*}
This indicates that the stationary point at $x = 4$ is a local minimum.