Question Statement

Find the root of the following equations using the bisection method correct up to two decimal places:

(a) $x^3-x^2+x-7=0$ on the interval $[2,3]$

(b) $x^3-2x-5=0$ on the interval $[2,3]$

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Background and Explanation

The bisection method is an iterative root-finding algorithm based on the Intermediate Value Theorem. It requires a continuous function $f(x)$ where $f(a)$ and $f(b)$ have opposite signs, guaranteeing at least one root in $[a,b]$. The method repeatedly halves the interval, selecting the subinterval where the sign change occurs, until the desired precision is achieved.

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Solution

Part (a): Solving $x^3-x^2+x-7=0$

Define the function: $f(x)=x^3-x^2+x-7$

Step 1: Verify the initial interval $[2,3]$

$\begin{array}{rl}f(2)& =2^3-2^2+2-7=8-4+2-7=-1<0\\ f(3)& =3^3-3^2+3-7=27-9+3-7=14>0\end{array}$

Since $f(2)<0$ and $f(3)>0$, a root lies between 2 and 3.

Step 2: First bisection of $[2,3]$

$\begin{array}{rl}{x}_0& =\frac{2+3}{2}=2.5\\ f\left(x_0\right)& =f(2.5)=(2.5{)}^3-(2.5{)}^2+2.5-7\\ & =15.625-6.25+2.5-7=4.875>0\end{array}$

Since $f(2.5)>0$ and $f(2)<0$, the updated interval is $[2,2.5]$.

Step 3: Second bisection of $[2,2.5]$

$\begin{array}{rl}{x}_1& =\frac{2+2.25}{2}=2.125\\ f\left(x_1\right)& =(2.125{)}^3-(2.125{)}^2+2.125-7\\ & =0.205>0\end{array}$

The updated interval is $[2,2.125]$.

Step 4: Third bisection of $[2,2.125]$

$\begin{array}{rl}{x}_2& =\frac{2+2.125}{2}=2.0625\\ f\left(x_2\right)& =f(2.0625)\\ & =(2.0625{)}^3-(2.0625{)}^2+2.0625-7\\ & =-0.417<0\end{array}$

The updated interval is $[2.0625,2.125]$.

Step 5: Fourth bisection of $[2.0625,2.125]$

$\begin{array}{rl}{x}_3& =\frac{2.0625+2.125}{2}=2.09375\\ f\left(x_3\right)& =f(2.09375)\\ & =(2.09375{)}^3-(2.09375{)}^2+2.09375-7\\ & =-0.1114<0\end{array}$

The updated interval is $[2.09375,2.125]$.

Step 6: Fifth bisection of $[2.09375,2.125]$

$\begin{array}{rl}{x}_4& =\frac{2.09375+2.125}{2}=2.109375\\ f\left(x_4\right)& =f(2.109375)\\ & =(2.109375{)}^3-(2.109375{)}^2+(2.109375)-7\\ & =0.045>0\end{array}$

The updated interval is $[2.09375,2.109375]$.

Step 7: Sixth bisection of $[2.09375,2.109375]$

$\begin{array}{rl}{x}_5& =\frac{2.09375+2.109375}{2}=2.1015625\\ f\left(x_5\right)& =f(2.1015625)\\ & =(2.1015625{)}^3-(2.1015625{)}^2+2.1015625-7\\ & =-0.033<0\end{array}$

The updated interval is $[2.1015625,2.109375]$.

Conclusion for Part (a): Up to two decimal places, the endpoints of the final interval are effectively the same (interval width $\approx 0.008<0.01$). Thus, correct up to two decimal places, the root of the equation is 2.10.

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Part (b): Solving $x^3-2x-5=0$

Define the function: $f(x)=x^3-2x-5$

Step 1: Verify the initial interval $[2,3]$

$\begin{array}{rl}f(2)& =2^3-2(2)-5=8-4-5=-1<0\\ f(3)& =3^3-2(3)-5=27-6-5=16>0\end{array}$

Root lies between 2 and 3.

Step 2: First bisection of $[2,3]$

$\begin{array}{rl}{x}_0& =\frac{2+3}{2}=2.5\\ f\left(x_0\right)& =f(2.5)=(2.5{)}^3-2(2.5)-5\\ & =15.625-5-5=5.625>0\end{array}$

The updated interval is $[2,2.5]$.

Step 3: Second bisection of $[2,2.5]$

$\begin{array}{rl}{x}_1& =\frac{2+2.5}{2}=2.125\\ f\left(x_1\right)& =f(2.25)\\ & =(2.25{)}^3-2(2.25)-5\\ & =11.390625-4.5-5=1.890>0\end{array}$

The updated interval is $[2,2.25]$.

Step 4: Third bisection of $[2,2.25]$

$\begin{array}{rl}{x}_2& =\frac{2+2.25}{2}=2.125\\ f\left(x_2\right)& =f(2.125)\\ & =(2.125{)}^3-2(2.125)-5\\ & =9.595703125-4.25-5=0.345>0\end{array}$

The updated interval is $[2,2.125]$.

Step 5: Fourth bisection of $[2,2.125]$

$\begin{array}{rl}{x}_3& =\frac{2+2.125}{2}=2.0625\\ f\left(x_3\right)& =f(2.0625)\\ & =(2.0625)-2(2.0625)-5\\ & =-0.35<0\end{array}$

The updated interval is $[2.0625,2.125]$.

Step 6: Fifth bisection of $[2.0625,2.125]$

$\begin{array}{rl}{x}_4& =\frac{2.0625+2.125}{2}=2.09375\\ f\left(x_4\right)& =f(2.09375)\\ & =(2.09375{)}^3-2(2.09375)-5\\ & =-0.0089<0\end{array}$

The updated interval is $[2.09375,2.125]$.

Step 7: Sixth bisection of $[2.09375,2.125]$

$\begin{array}{rl}{x}_5& =\frac{2.09375+2.125}{2}=2.109375\\ f\left(x_5\right)& =f(2.109375)\\ & =(2.109375)-2(2.109375)-5\\ & =0.166>0\end{array}$

The updated interval is $[2.09375,2.109375]$.

Step 8: Seventh bisection of $[2.09375,2.109375]$

$\begin{array}{rl}{x}_6& =\frac{2.09375+2.109375}{2}=2.1015625\\ f\left(x_6\right)& =f(2.1015625)\\ & =(2.1015625{)}^3-2(2.1015625)-5\\ & =0.078>0\end{array}$

The updated interval is $[2.09375,2.1015625]$.

Step 9: Eighth bisection of $[2.09375,2.1015625]$

$\begin{array}{rl}{x}_7& =\frac{2.09375+2.1015625}{2}=2.09765625\\ f\left(x_7\right)& =f(2.09765625)\\ & =(2.09765625)-2(2.09765625)-5\\ & =0.0347>0\end{array}$

The updated interval is $[2.09375,2.09765625]$.

Conclusion for Part (b): Up to two decimal places, the endpoints of the final interval are the same (interval width $\approx 0.004<0.01$). Hence, correct up to two decimal places, the root of the equation is 2.10.

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Key Formulas or Methods Used

• Intermediate Value Theorem: If $f$ is continuous on $[a,b]$ and $f(a)\cdot f(b)<0$, then there exists $c\in (a,b)$ such that $f(c)=0$
• Bisection Formula: $x_n=\frac{a_n+b_n}{2}$
• Interval Update Rule:
  • If $f(x_n)\cdot f(a_n)<0$, new interval is $[a_n,x_n]$
  • If $f(x_n)\cdot f(b_n)<0$, new interval is $[x_n,b_n]$
• Stopping Criterion: When interval width $<0.01$ (for two decimal place accuracy)

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Summary of Steps

1. Verify bracketing: Check that $f(a)$ and $f(b)$ have opposite signs
2. Compute midpoint: Calculate $x=\frac{a+b}{2}$
3. Evaluate function: Find $f(x)$ at the midpoint
4. Update interval: Select the subinterval where the sign change occurs
5. Check precision: If interval width $<0.01$, stop; otherwise return to step 2
6. Report result: The midpoint of the final interval (or either endpoint) gives the root correct to two decimal places