Stationary Points & the Second Derivative

A stationary point is where the gradient vanishes, $\frac{dy}{dx} = 0$, and 0606 examines the full routine in essentially every session: find, classify, conclude.

The routine

Find the stationary points of $y = 2x^3 - 9x^2 + 12x - 1$ and determine their nature. $\frac{dy}{dx} = 6x^2 - 18x + 12$ (M, A) Set $\frac{dy}{dx} = 0$ (write this line, it is routinely a mark by itself): $6(x - 1)(x - 2) = 0 \to x = 1, 2$ Points: $(1, 4)$ and $(2, 3)$ ($y$-coordinates computed. “points” means both coordinates) $\frac{d^2y}{dx^2} = 12x - 18$ At $x = 1$: $-6 < 0 \to$ $(1, 4)$ is a maximum; at $x = 2$: $+6 > 0 \to$ $(2, 3)$ is a minimum

The classification sentences must show the second-derivative value and sign and state the verdict in words. ”$\frac{d^2y}{dx^2} = -6 < 0$, hence maximum”. A bare “max” next to a number drops the conclusion mark. If $\frac{d^2y}{dx^2} = 0$, the test is inconclusive: fall back on checking the gradient’s sign either side.

Applied maximum/minimum, the modelling layer

The heavyweight version wraps the routine in a story:

“A box with square base $x$ and volume 32 has surface area $S = x^2 + \frac{128}{x}$. Find the value of $x$ that minimises $S$.” $\frac{dS}{dx} = 2x - \frac{128}{x^2} = 0 \to x^3 = 64 \to x = 4$; $\frac{d^2S}{dx^2} = 2 + \frac{256}{x^3} > 0 \to$ minimum ✓

When the formula isn’t given, building it is the first marks: write the constraint (volume $= 32$), express one variable in terms of the other, substitute into the target quantity. Then the routine runs as above, and the final sentence answers the actual question (“the minimum surface area is $S(4) = 48$”, if the area was asked rather than $x$). Real-context answers also obey reality: negative lengths get rejected, in writing.

Increasing and decreasing

The same derivative classifies behaviour between stationary points: $\frac{dy}{dx} > 0 \to$ increasing. “Find the range of $x$ for which $y$ is decreasing” $\to$ solve $\frac{dy}{dx} < 0$, a quadratic inequality. The stationary points are the boundaries, find them first and the regions follow.

Common mistakes

• "$\frac{dy}{dx} = 0$" used but never written
• $y$-coordinates omitted when points were asked
• Classification verdicts unjustified or unworded
• Applied questions answered with $x$ when the quantity was asked (or vice versa)
• The inconclusive $\frac{d^2y}{dx^2} = 0$ case force-classified

Full topic context: Calculus notes.