A curve C has equation y = x e^(2x). (a) Find dy/dx. (b) Find the exact coordinates of the stationary point of C. (c) Determine the nature of this stationary point.

Mark scheme

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

1. M1: Uses the product rule correctly, u=x, v=e^(2x) leading to du/dx=1, dv/dx=2e^(2x) or equivalent.
2. A1: dy/dx = e^(2x) + 2x e^(2x) or e^(2x)(1+2x).
3. M1: Sets dy/dx = 0 and attempts to solve for x. E.g., e^(2x)(1+2x) = 0.
4. A1: For x = -1/2 correctly found. (e^(2x) ≠ 0).
5. A1: For y = -1/2 e^(-1) or -1/(2e) correctly found. So, (-1/2, -1/(2e)).
6. M1: Uses the product rule again (or chain rule on e^(2x) and then product rule) to find d²y/dx².
7. A1: For d²y/dx² = 2e^(2x)(1+2x) + e^(2x)(2) = 4e^(2x) + 4x e^(2x) or 4e^(2x)(1+x).
8. A1: Substitutes x = -1/2 into d²y/dx² to get d²y/dx² = 4e^(-1)(1-1/2) = 2e^(-1) or 2/e. Since this is > 0, states that it is a minimum point.