Stationary Points and the Second Derivative Test

A curve’s stationary points are where $\frac{dy}{dx} = 0$. To classify each as a maximum or minimum, the cleanest method is the second derivative test: the sign of $\frac{d^2y}{dx^2}$ tells you the concavity, and hence the nature.

The curve $y = x^3 - 3x^2 + 4$ has two stationary points. Find their coordinates and determine the nature of each. [6]

The working

Step 1, differentiate and set to zero: $\frac{dy}{dx} = 3x^2 - 6x = 3x(x - 2) \quad \text{(M1)}$ $3x(x - 2) = 0 \implies x = 0 \ \text{or} \ x = 2 \quad \text{(A1)}$

Step 2, find the $y$-coordinates:

• $x = 0$: $\ y = 0 - 0 + 4 = 4$, giving $(0, 4)$.
• $x = 2$: $\ y = 8 - 12 + 4 = 0$, giving $(2, 0)$. (A1)

Step 3, second derivative: $\frac{d^2y}{dx^2} = 6x - 6 \quad \text{(M1)}$

Step 4, test the sign at each point:

• At $x = 0$: $\ 6(0) - 6 = -6 < 0$, so $(0, 4)$ is a maximum. (A1)
• At $x = 2$: $\ 6(2) - 6 = 6 > 0$, so $(2, 0)$ is a minimum. (A1)

Where the marks are won and lost

• You must state the conclusion (“maximum” / “minimum”), not just the sign of the second derivative. The sign is the evidence; the classification is the answer.
• Give coordinates, both $x$ and $y$. A common slip is to find the $x$-values and stop.
• $\frac{d^2y}{dx^2} < 0$ is the maximum (concave down, like a hill). Students often flip this, so anchor it: negative curvature curves downward, which is a peak.

Common mistakes

• Reading the sign backwards (negative $\to$ minimum).
• Forgetting to substitute back for the $y$-coordinates.
• Solving $\frac{d^2y}{dx^2} = 0$ instead of $\frac{dy}{dx} = 0$ for the stationary points.

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