Paper 2 · June 2012 · Numerical methods, Newton-Raphson

Use the Newton-Raphson method to find, correct to 3 decimal places, the root of the equation $\sin x+\cos x=e^{x}-1$ that lies between $-4$ and $-3$ .

Model answer

-3,182

Also accepted: -3.182, x=-3,182, x = -3.182

Explanation

Write

f(x)=\sin x+\cos x-e^{x}+1

, so

f'(x)=\cos x-\sin x-e^{x}

. Starting from

x_0=-3

, the iteration

x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}

gives

x_1=-3,201299

x_2=-3,182417

x_3=-3,182268

and

x_4=-3,182268

. The root is

-3,182

to 3 decimal places.

Derived from ZIMSEC Mathematics Paper 2, June 2012, Q1

More questions from this paper

Danho has every sitting for this paper, with your progress tracked question by question, offline.