Question 6 of 7: Fixed-Point Iteration, Bisection/Newton, and Extrema of a Quartic

Question 6: Fixed-Point Iteration, Bisection/Newton, and Extrema of a Quartic (a: 7; b: 7; c: 6 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. $f(x)=x^4-3x^3+5x-6$. A root near $x_0=2$; another root bracketed by $[-2,-1]$; and the fact that $x=1$ is a root of $f'(x)=0$.

Find. (a) Six fixed-point iterates converging toward the root near $2$. (b) Three bisection steps on $[-2,-1]$ followed by three Newton steps. (c) The remaining two critical points, their classification, and a sketch of $f(x)$.

Approach. (a) Rearrange the quartic into a form $x=g(x)$ with $|g'|\lt1$ near $x_0=2$ so the iteration converges. (b) Halve the given bracket three times, then hand the resulting midpoint to Newton–Raphson for three quadratically-convergent steps. (c) Factor $f'(x)$ using the given root $x=1$, solve the remaining quadratic, and classify all three critical points with the second-derivative test.

1. (a) Choose a convergent rearrangement. Isolating the $x^4$ and $-3x^3$ terms together, $x^3(x-3)=6-5x\Rightarrow x-3=\dfrac{6-5x}{x^3}$, i.e. $$\boxed{g(x)=3-\dfrac{5x-6}{x^{3}}}$$ Near $x=2$, $|g'(2)|\approx0.4\lt1$, so this rearrangement converges — unlike, e.g., isolating $x=[(3x^3-5x+6)]^{1/4}$, which converges much more slowly from the same start.
2. Iterate six times from $x_0=2$. $$x_0=2.0000000,\ x_1=2.5000000,\ x_2=2.5840000,\ x_3=2.5989223,\ x_4=2.6015409,\ x_5=2.6019993,\ x_6=\boxed{2.6020795}$$ (The exact root is $x^\ast\approx2.6020966$, so six iterations already agree to five significant figures.)
3. (b) Three bisections on $[-2,-1]$. $f(-2)=24,\ f(-1.5)=1.6875,\ f(-1.25)=-3.949219,\ f(-1)=-7$ (opposite signs of $f(-2)$ and $f(-1)$ confirm a root is bracketed).
   Step 1: midpoint $-1.5$; $f(-2)$ and $f(-1.5)$ share sign $(+,+)$, so the root lies in $(-1.5,-1)$.
   Step 2: midpoint $-1.25$; $f(-1.5)=+1.6875$, $f(-1.25)=-3.949219$ — sign change, so the root lies in $(-1.5,-1.25)$.
   Step 3: midpoint $-1.375$; $f(-1.375)=-1.501709$, opposite sign to $f(-1.5)=+1.6875$, so the root lies in $(-1.5,-1.375)$.
   Best bracket estimate after three bisections: $\boxed{x\approx-1.4375000}$ (the bracket midpoint).
4. Three Newton steps from $x_0=-1.4375000$. With $f'(x)=4x^3-9x^2+5$, $$x_1=x_0-\dfrac{f(x_0)}{f'(x_0)}=-1.4375000-\dfrac{-6.088\times10^{-3}}{4.284}=\boxed{-1.4377389}$$ $$x_2=-1.4377389,\qquad x_3=-1.4377389$$ Newton's quadratic convergence already reproduces the root $x^\ast\approx-1.4377389$ to seven digits after a single step from the bisection-refined start.
5. (c) Differentiate and factor. $f'(x)=4x^3-9x^2+5$. Since $f'(1)=4-9+5=0$, $(x-1)$ is a factor; dividing, $$f'(x)=(x-1)(4x^2-5x-5)$$ Solving $4x^2-5x-5=0$ by the quadratic formula, $x=\dfrac{5\pm\sqrt{25+80}}{8}=\dfrac{5\pm\sqrt{105}}{8}$, giving $$\boxed{x\approx-0.6558688\quad\text{and}\quad x\approx1.9058688}$$
6. Classify the three critical points. $f''(x)=12x^2-18x$. $$f''(-0.6558688)\approx16.97\gt0\ (\text{local min}),\quad f''(1)=-6\lt0\ (\text{local max}),\quad f''(1.9058688)\approx9.28\gt0\ (\text{local min})$$ with function values $$f(-0.6558688)=\boxed{-8.247910}\ (\text{min}),\quad f(1)=\boxed{-3}\ (\text{max}),\quad f(1.9058688)=\boxed{-4.045059}\ (\text{min})$$ Because the local maximum value $(-3)$ is still negative, $f(x)$ never rises above the $x$-axis between the two minima — consistent with $f$ having exactly the two real roots found in (a) and (b) ($x\approx2.6021$ to the right of the rightmost minimum, and $x\approx-1.4377$ to the left of the leftmost minimum), with the other two roots of the quartic complex.