Key Idea: A stationary point is where the gradient equals zero — the curve momentarily flattens out. It could be a local maximum (peak), a local minimum (trough), or a point of inflection (S-curve). Topic 5.6 is about finding these points, classifying them, and determining the intervals where the function is increasing or decreasing.
✅ Finding and classifying stationary points
Example: f(x) = x³ − 6x² + 9x + 1 f'(x) = 3x² − 12x + 9 = 3(x−1)(x−3) Stationary points: x = 1 and x = 3. f(1) = 1 − 6 + 9 + 1 = 5 → point (1, 5) f(3) = 27 − 54 + 27 + 1 = 1 → point (3, 1) f''(x) = 6x − 12. f''(1) = −6 < 0 → (1, 5) is a local maximum f''(3) = 6 > 0 → (3, 1) is a local minimum Increasing: x < 1 and x > 3 (f'(x) > 0 in those intervals) Decreasing: 1 < x < 3 (f'(x) < 0)
f''(x) = 0 is inconclusive — it does NOT prove a point of inflection. Use a sign chart on f'(x) to check whether the gradient actually changes sign around that x. For increasing/decreasing: solve f'(x) > 0 and f'(x) < 0 as inequalities, or sketch f'(x) and read off where it is above/below the x-axis.
Paper 1 (GDC allowed): Show f'(x) = 0 → factorisation → x values → y values → f''(x) classification. Each step is a marking point. Paper 2 (GDC allowed): Use GDC 'minimum' and 'maximum' functions to verify coordinates. But still show the differentiation and second derivative test in your written working.
IB-style question [7 marks]
The function f is defined by f(x) = 2x³ − 3x² − 12x + 5. (a) Find f′(x). (b) Find the coordinates of the stationary points of f. (c) Classify each stationary point as a local maximum or a local minimum.
Step by step:
(a) Differentiate term by term.
(b) Set f′(x) = 0 and factorise to find the stationary x-values.
Find the y-coordinates by substituting back into f.
(c) Graph y = f(x) on the GDC: the left stationary point is a peak, the right one is a valley.
Final answer:
(a) f′(x) = 6x² − 6x − 12. (b) (−1, 12) and (2, −15). (c) (−1, 12) is a local maximum; (2, −15) is a local minimum.