The equation e^(x) + 2x - 3 = 0 has a single root α. (a) Show that α lies between 0.5 and 0.6. (b) Using x₀ = 0.55 as a first approximation, apply the Newton-Raphson method once to find a second approximation to α, giving your answer to 4 decimal places.

Mark scheme

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

1. M1: Defines f(x) = e^x + 2x - 3 and evaluates f(0.5) and f(0.6).
2. A1: f(0.5) = e^0.5 + 2(0.5) - 3 = 1.6487 + 1 - 3 = -0.35... and f(0.6) = e^0.6 + 2(0.6) - 3 = 1.8221 + 1.2 - 3 = 0.022... and states 'sign change, hence root between 0.5 and 0.6'.
3. B1: States the Newton-Raphson formula x_(n+1) = x_n - f(x_n)/f'(x_n).
4. M1: Finds f'(x) = e^x + 2.
5. M1: Evaluates f(0.55) = e^0.55 + 2(0.55) - 3 = 1.73325 + 1.1 - 3 = -0.16675...
6. A1: Evaluates f'(0.55) = e^0.55 + 2 = 1.73325 + 2 = 3.73325...
7. A1: Substitutes into the formula: x_1 = 0.55 - (-0.16675 / 3.73325) = 0.55 + 0.04466... = 0.5947 (4 d.p.).