Question 6 of 7: Bisection, Newton–Raphson and Fixed-Point Iteration

Question 6: Bisection, Newton–Raphson and Fixed-Point Iteration (20 marks: A(a) 5, A(b) 7, B 8)

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. Part (A): $f(x)=x^4-3x^2+5x-12$ on the bracket $[1,2]$. Part (B): $h(x)=\ln(x+5)-x^2+x$ with a root near $x_0=2.0$. Seven significant digits are to be carried throughout part A(b) and part B.

Find. A(a) three bisection steps; A(b) two Newton–Raphson steps started from the A(a) result; B a fixed-point form $x=g(x)$ and six iterations from $x_0=2.0$.

Approach. Confirm the sign change first, then halve the bracket three times; switch to Newton–Raphson for quadratic convergence once the estimate is close. For part B, isolate $x$ from the $-x^2$ term to build a contraction mapping and iterate.

1. A(a) — confirm the bracket. Evaluating the endpoints, $$f(1)=1-3+5-12=-9,\qquad f(2)=16-12+10-12=+2$$ The sign change confirms at least one root in $(1,2)$, so bisection is guaranteed to converge.
2. A(a) — three bisections. Each step takes the midpoint and keeps the half that still brackets a sign change: $$c_1=\tfrac{1+2}{2}=1.500000,\quad f(c_1)=-6.187500\ (\lt0)\ \Rightarrow\ \text{root in }[1.5,\,2]$$ $$c_2=\tfrac{1.5+2}{2}=1.750000,\quad f(c_2)=-3.058594\ (\lt0)\ \Rightarrow\ \text{root in }[1.75,\,2]$$ $$c_3=\tfrac{1.75+2}{2}=1.875000,\quad f(c_3)=-0.812256\ (\lt0)\ \Rightarrow\ \text{root in }[1.875,\,2]$$ After three bisections the approximation is $$\boxed{x\approx c_3=1.875000}$$ with the root known to lie in an interval of width $2-1.875=0.125$.

3. A(b) — Newton–Raphson, first step. The derivative is $f'(x)=4x^3-6x+5$. Starting from $x_0=1.875000$, $$f(1.875000)=-0.8122559,\qquad f'(1.875000)=4(6.591797)-6(1.875)+5=20.11719$$ $$x_1=x_0-\dfrac{f(x_0)}{f'(x_0)}=1.875000-\dfrac{-0.8122559}{20.11719}=1.9153762$$
4. A(b) — second step. Repeating at $x_1$, $$f(1.9153762)=+0.0299935,\qquad f'(1.9153762)=21.61525$$ $$\boxed{x_2=1.9153762-\dfrac{0.0299935}{21.61525}=1.9139886}$$ Against the true root $1.9139869$ this is accurate to six decimal places. The contrast with part (a) is the point of the question: three bisections cut the error to about $4\times10^{-2}$, whereas two Newton steps from that same point cut it to about $2\times10^{-6}$ — the error roughly squares at each Newton step because the method is quadratically convergent near a simple root.
5. B — construct the iteration function. Rearranging $\ln(x+5)-x^2+x=0$ to isolate the quadratic term gives $x^2=\ln(x+5)+x$, and taking the positive square root (the root sought is near $+2$) yields $$x=g(x)=\sqrt{\ln(x+5)+x}$$ This form converges because $|g'(x)|\lt1$ near the root: differentiating, $g'(x)=\dfrac{1+\dfrac{1}{x+5}}{2\sqrt{\ln(x+5)+x}}$, and at $x\approx1.98$ this evaluates to $0.2886$, comfortably inside the unit interval. (Solving instead for $x$ from the linear term, $x=x^2-\ln(x+5)$, gives $|g'|\approx3.8\gt1$ and diverges — the choice of rearrangement is not arbitrary.)
6. B — six iterations from $x_0=2.0$. Carrying seven digits throughout:

   $n$	$x_n$	$x_{n+1}=g(x_n)$
   1	2.0000000	1.9864315
   2	1.9864315	1.9825240
   3	1.9825240	1.9813971
   4	1.9813971	1.9810720
   5	1.9810720	1.9809782
   6	1.9809782	1.9809511

   After six iterations, $$\boxed{x\approx1.9809511}$$ The true root is $1.9809401$, so the sixth iterate is accurate to about $1.1\times10^{-5}$. Successive corrections shrink by a factor of roughly $0.29$ each time, exactly the value of $|g'|$ computed in Step 5 — the signature of linear convergence, and the reason this method needs six steps to reach an accuracy Newton–Raphson would reach in two.