Question 6 of 7: Root-Finding by Bisection, Newton's Method, and Fixed-Point Iteration

Question 6: Root-Finding by Bisection, Newton's Method, and Fixed-Point Iteration (5+8+7 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given (A)/(B). $f(x)=6^x-13x$; $f(1)=-7$, $f(2)=10$ (a sign change, so a root lies in $[1,2]$).

Find (A). Four bisection midpoints. Find (B). Three further Newton iterates starting from the fourth bisection midpoint.

Approach. Bisection: repeatedly halve the sign-changing bracket. Newton: $x_{n+1}=x_n-f(x_n)/f'(x_n)$ with $f'(x)=6^x\ln6-13$.

1. Bisection iteration 1. $c_1=(1+2)/2=1.5$; $f(1.5)=6^{1.5}-19.5=14.697-19.5=-4.803$ (same sign as $f(1)$) → new bracket $[1.5,2]$.
2. Bisection iteration 2. $c_2=1.75$; $f(1.75)=6^{1.75}-22.75=22.998-22.75=+0.248$ → bracket $[1.5,1.75]$.
3. Bisection iteration 3. $c_3=1.625$; $f(1.625)=6^{1.625}-21.125=18.379-21.125=-2.746$ → bracket $[1.625,1.75]$.
4. Bisection iteration 4. $c_4=1.6875$; $f(1.6875)=6^{1.6875}-21.9375=20.578-21.9375=-1.360$ → bracket $[1.6875,1.75]$. $$\boxed{c_4=1.6875}$$
5. Newton, starting from $x_0=c_4=1.6875000$. $f'(x)=6^x\ln6-13$; $f'(1.6875)=20.578(1.791759)-13=23.865$. $$x_1=1.6875000-\frac{-1.360}{23.865}=1.7450498$$
6. Newton iterations 2 and 3. Repeating the update (all values carried to 7 digits): $$x_2=1.7409856,\qquad x_3=\boxed{1.7409637}$$ (three Newton steps already match the true root $1.7409637\ldots$ to 7 digits — the expected quadratic convergence.)

Given (C). $h(x)=\ln(x+5)-x^2+5x+6=0$, rearranged as $x=g(x)=0.2\big[x^2-6-\ln(x+5)\big]$, $x_0=-1$.

Find (C). Six fixed-point iterates $x_1,\dots,x_6$, and a justification of convergence.

Approach (C). Iterate $x_{k+1}=g(x_k)$ six times, then check the convergence criterion $|g'(x)|<1$ near the root.

1. Iterate $x_{k+1}=g(x_k)$ six times from $x_0=-1.000000$ (6 digits). $$x_1=-1.277259,\ x_2=-1.136614,\ x_3=-1.211931,\ x_4=-1.172616,\ x_5=-1.193431,\ x_6=\boxed{-1.182490}$$ (the true root is $-1.186273\ldots$; the iterates oscillate in and converge toward it.)
2. Justify convergence. $g'(x)=0.2\big[2x-\tfrac1{x+5}\big]$. Near the root, $g'(-1.186)\approx-0.527$, so $|g'(x)|\approx0.53<1$ throughout the region visited — the fixed-point iteration theorem (Kreyszig) guarantees linear convergence to the root for this $g$, consistent with the oscillating-but-shrinking iterates above.