Q.2 (b) Using the Newton-Raphson method find the root of the equation \(2x^3-3x+4=0\) lying between -2 and -1, correct to four places of decimals.
Best answer
\(f(x)=2x^3-3x+4,\ f'(x)=6x^2-3\). \(f(-2)=-6,\ f(-1)=5\) → root in (−2, −1); \(f(-1.5)=1.75\), so take \(x_0=-1.5\).
\(x_{n+1}=x_n-\frac{2x_n^3-3x_n+4}{6x_n^2-3}=\frac{4x_n^3-4}{6x_n^2-3}\)
\(x_1=\frac{4(-3.375)-4}{6(2.25)-3}=\frac{-17.5}{10.5}=-1.666667\)
\(x_2=\frac{4(-4.629630)-4}{6(2.777778)-3}=\frac{-22.518519}{13.666667}=-1.647696\)
\(x_3=-1.647427\)
\(x_4=-1.647427\)
Root \(\approx-1.6474\) (check: \(f(-1.6474)\approx0.0005\approx0\)).
Log in or create a free account to answer this question.
