4. Stationary Points & Inflection

Video coming soon2 worked examples

Theory

Stationary points occur where $f'(x) = 0$ . Classify them using the second derivative: $f''(x) > 0$ gives a minimum, $f''(x) < 0$ gives a maximum. A point of inflection is where the concavity changes — where $f''(x) = 0$ and $f''$ changes sign.

The Golden Rule: solve $f'(x) = 0$ for the stationary points, then use the sign of $f''(x)$ to classify each; for an inflection, confirm the concavity actually changes.

⚠️ Common Examiner Traps

$f''(x) = 0$ is not enough: an inflection also needs a genuine change of sign in $f''$ .
Missing the $y$ -coordinate: give full coordinates for each stationary point.
Nature confusion: keep the sign test straight — positive second derivative is a minimum.

Worked examples

Example 1

Find and classify the stationary points of $y = x^3 - 3x$ .

Step 1: Set $\frac{dy}{dx} = 0$ :

3x^2 - 3 = 0 \implies x = \pm 1

Step 2: Use $\frac{d^2y}{dx^2} = 6x$ to classify:

x = 1:\ 6 > 0 \ (\text{min}), \qquad x = -1:\ -6 < 0 \ (\text{max})

Step 3: Find the $y$ -coordinates: minimum at $(1, -2)$ , maximum at $(-1, 2)$ .

Example 2

Find the point of inflection of $y = x^3 - 6x^2 + 5$ .

Step 1: Differentiate twice and set the second derivative to zero:

\frac{dy}{dx} = 3x^2 - 12x, \qquad \frac{d^2y}{dx^2} = 6x - 12 = 0 \implies x = 2

Step 2: Check the sign change: $f''(x) < 0$ for $x < 2$ and $f''(x) > 0$ for $x > 2$ , so the concavity changes.

Step 3: Evaluate $y$ : $8 - 24 + 5 = -11$ . The point of inflection is $(2, -11)$ .