A curve has equation `y = x^3 - 3x^2 - 9x + 1`. (a) Find `dy/dx`. (b) Find the exact coordinates of the stationary points of the curve. (c) Determine the nature of each stationary point, giving reasons for your answers.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. M1 for `x^n` to `nx^(n-1)` for at least one term.
2. A1 for `dy/dx = 3x^2 - 6x - 9`.
3. M1 for setting `dy/dx = 0`.
4. A1 for `3(x^2 - 2x - 3) = 0 \Rightarrow 3(x-3)(x+1) = 0 \Rightarrow x=3` or `x=-1`.
5. M1 for substituting at least one of their `x` values into `y = x^3 - 3x^2 - 9x + 1`.
6. A1 for `(3, -26)` and `(-1, 6)`.
7. M1 for finding `d^2y/dx^2`. (`d^2y/dx^2 = 6x-6`).
8. A1 for `d^2y/dx^2` at `x=3` gives `12` (or `>0` or positive) AND `d^2y/dx^2` at `x=-1` gives `-12` (or `<0` or negative).
9. A1 for `(3, -26)` is a minimum, `(-1, 6)` is a maximum.
10. C1 for stating a valid reason for each, e.g. `d^2y/dx^2 > 0` for minimum, `d^2y/dx^2 < 0` for maximum. (Must link the nature to their value of the second derivative). (Alternative: using gradients before/after, or change of sign of `dy/dx`).