Solve f'(x) = 0: A stationary point is where the gradient is zero. Differentiate and solve f'(x) = 0 to find the x-values; a cubic typically has two stationary points (a maximum and a minimum).
[Diagram: math-stationary-points] - Available in full study mode
IB-style question — find them
Find the x-coordinates of the stationary points of f(x) = x³ − 6x² + 9x.
Step by step
- Differentiate and set to 0.
- Solve.
Final answer
Stationary points at x = 1 and x = 3.
Factor the derivative: Factor out common factors first (here 3), then factorise — it makes solving f'(x) = 0 quick.
Second-derivative test (or sign of f'): Classify each stationary point: with the second derivative, f''(x) > 0 → minimum, f''(x) < 0 → maximum. (If f'' = 0, check the sign of f' just before and after instead.)
IB-style question — classify
Classify the stationary points of f(x) = x³ − 6x² + 9x (at x = 1 and x = 3).
Step by step
- Second derivative.
- Test each.
Final answer
Maximum at x = 1, minimum at x = 3.
Min holds water: f'' > 0 (concave up, ∪) is a minimum; f'' < 0 (concave down, ∩) is a maximum.
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Substitute x back into f for the y-coordinate: For full coordinates, substitute each stationary x-value into the original f(x) to get y. A complete answer gives the point and its nature.
IB-style question — coordinates and nature
Find and classify the stationary points of f(x) = x³ − 6x² + 9x, giving coordinates.
Step by step
- From x = 1 (max) and x = 3 (min), find y.
- State the points.
Final answer
Maximum (1, 4); minimum (3, 0).
y comes from f, not f': Use the original function for the y-coordinate — substituting into f'(x) gives 0, not y.