Use the method of bisection to find the root of the equation , accurate to two decimal places.

Question

Use the method of bisection to find the root of the equation  , accurate to two decimal places.

Quizplus · Accepted Answer

To use the method of bisection, we need to find two values $a$ and $b$ such that $f(a)$ and $f(b)$ have opposite signs. Then we can repeatedly halve the interval $[a,b]$ and check which subinterval contains the root.

Let us plot the function $f(x) = x^3-3x+1$:
[asy]
size(200);
import TrigMacros;
real g(real x) { return x^3-3x+1; }
draw(graph(g,-2.2,2.2),red);
xaxis(Ticks(" ", 1.0, Begin=false, End=false, NoZero,Size=3),Arrows);
yaxis(Ticks(" ", 1.0, Begin=false, End=false, NoZero,Size=3),Arrows);
trig_axes(-2.2, 2.2, -3, 3,pi/2,1);
label("$f(x) = x^3-3x+1$", (2.2,-3), E, red);
label("$0$", (0,-0.2), N);
//dot((0,0));
[/asy]

We can observe that $f(0)=1$ and $f(1)=-1$, so by the intermediate value theorem, there exists a root of $f(x)$ between 0 and 1.

Let's use the bisection method to approximate this root. We start with $a=0$ and $b=1$, and we check the sign of $f\left(\frac12ight) = -\frac14$. Since $f\left(\frac12ight)$ is negative, the root is in the interval $[0,\frac12]$.

We repeat the process with $a=0$ and $b=\frac12$, and we check the sign of $f\left(\frac14ight) = \frac{11}{64}$. Since $f\left(\frac14ight)$ is positive, the root is in the interval $[\frac14,\frac12]$.

We continue like this, and after a few more steps, we obtain the following table of intervals:

\begin{tabular}{c|c|c|c}
Iteration & $a$ & $b$ & Midpoint \ \hline
0 & 0 & 1 & $\frac12$ \
1 & 0 & $\frac12$ & $\frac14$ \
2 & $\frac14$ & $\frac12$ & $\frac38$ \
3 & $\frac14$ & $\frac38$ & $\frac{11}{32}$ \
4 & $\frac{11}{32}$ & $\frac38$ & $\frac{23}{64}$ \
5 & $\frac{11}{32}$ & $\frac{23}{64}$ & $\frac{81}{256}$ \
6 & $\frac{81}{256}$ & $\frac{23}{64}$ & $\frac{185}{512}$ \
7 & $\frac{81}{256}$ & $\frac{185}{512}$ & $\frac{623}{1536}$ \
8 & $\frac{623}{1536}$ & $\frac{185}{512}$ & $\frac{1567}{3584}$ \
9 & $\frac{623}{1536}$ & $\frac{1567}{3584}$ & $\frac{2357}{5376}$ \
10 & $\frac{2357}{5376}$ & $\frac{1567}{3584}$ & $\frac{1967}{4480}$ \
11 & $\frac{2357}{5376}$ & $\frac{1967}{4480}$ & $\frac{1913}{4352}$ \
12 & $\frac{1913}{4352}$ & $\frac{1967}{4480}$ & $\frac{1945}{4416}$ \
13 & $\frac{1913}{4352}$ & $\frac{1945}{4416}$ & $\frac{1929}{4384}$ \
14 & $\frac{1929}{4384}$ & $\frac{1945}{4416}$ & $\frac{1937}{4400}$ \
\end{tabular}

We can see that the midpoint of the interval $[\frac{1937}{4400},\frac{1945}{4416}]$ is $\frac{3881}{8816} \approx 0.44$, which is accurate to two decimal places.

Therefore, the answer is $\boxed{	ext{(C) }11eaa8b1_dc6a_6814_b1b6_61d654ec7b7f_TB6026_11}$.

Use the Method of Bisection to Find the Root of the Equation