Question 6 of 7: Root-Finding — Newton, Fixed-Point Iteration, and Bisection
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Examination — 04-BS-5 Advanced Mathematics, December 2015. Closed-book, 3-hour exam;
candidates were permitted one 8.5"x11" aid sheet (both sides) and an approved calculator. The exam
instructs that any five questions constitute a complete paper (first five answers in the answer book are marked); every question is solved below as a full study resource.
Given. (A)/(B) $f(x)=\ln(x+2)-x^2+6x-5$, which has two real roots in the region examined here
(near $x\approx5.45$ and $x\approx0.76$); (C) $g(x)=2\cos(x/2)-x-1$, bracketed by $[0.82,0.84]$.
Find. (A) Three Newton iterates from $x_0=5.0$; (B) six fixed-point iterates from $x_0=1$
using the given $g(x)$, plus a convergence justification; (C) three bisection midpoints from $[0.82,0.84]$.
Approach. Apply each method's standard recursion directly to the given starting data, carrying
seven significant digits as instructed, and check convergence via $|g'(\text{root})|<1$ for part (B).
(A) Newton's method, $x_{k+1}=x_k-f(x_k)/f'(x_k)$. With $f'(x)=\dfrac{1}{x+2}-2x+6$ and
$x_0=5.0$: $f(5)=\ln7=1.945910$, $f'(5)=\tfrac17-4=-3.857143$, so
$$x_1 = 5.0 - \frac{1.945910}{-3.857143} = 5.504495.$$
Repeating twice more with seven-digit arithmetic:
$$x_2 = 5.451786, \qquad x_3 = \boxed{5.451199}.$$
(The iteration overshoots past $x_1$ then settles back — $f''$ is non-negligible near $x_0$, so the first
Newton step is not yet in the quadratically-convergent regime; by $x_3$ successive digits have stabilized.)
(B) Fixed-point iteration, $x_{k+1}=g(x_k)=\dfrac{\ln(x_k+2)-5}{x_k-6}$, from $x_0=1$.
$$x_1=g(1)=\frac{\ln3-5}{1-6}=\frac{-3.901388}{-5}=0.7802775.$$
Continuing five more times:
$$x_2=0.7620040,\ x_3=0.7606045,\ x_4=0.7604981,\ x_5=0.7604900,\ x_6=\boxed{0.7604894}.$$
This converges to the second real root of $f(x)=0$ near $x\approx0.7605$ (a root the exam's part (A)
setup did not target, since Newton from $x_0=5$ stays near the upper root).
(B) Convergence justification. Fixed-point iteration $x_{k+1}=g(x_k)$ converges to a root
$x^\ast$ whenever $|g'(x^\ast)|\lt 1$ in a neighbourhood of $x^\ast$ (contraction mapping principle). Differentiating
$g(x)=\dfrac{\ln(x+2)-5}{x-6}$ and evaluating at the converged root $x^\ast\approx0.760489$ gives
$$g'(x^\ast) \approx 0.0760 ,$$
and $|0.0760|\ll 1$, so the iteration is a strong contraction there — consistent with the fast settling seen
after only 3–4 iterations. (By contrast, solving the SAME equation for $x=g(x)$ in the form
$x=\sqrt{6x-5+\ln(x+2)}$ near this root would give $|g'|$ close to or above 1 and could diverge — the specific
algebraic rearrangement given in the question was chosen precisely because it is a contraction near this root.)
(C) Bisection, $[\alpha,\beta]=[0.82,0.84]$. With $h(x)=2\cos(x/2)-x-1$: $h(0.82)=+0.01424\gt 0$,
$h(0.84)=-0.01382\lt 0$, confirming the sign change. Each step bisects and keeps the half with the sign change:
$$m_1=\frac{0.82+0.84}{2}=0.830 \ (h(m_1)\gt 0\Rightarrow\text{root in }[0.83,0.84]),$$
$$m_2=\frac{0.83+0.84}{2}=0.835 \ (h(m_2)\lt 0\Rightarrow\text{root in }[0.83,0.835]),$$
$$m_3=\frac{0.83+0.835}{2}=\boxed{0.8325}.$$
