Question Statement

Find the root of the equation $x^3-4x-1=0$ in the interval $[2,3]$ using the Regula Falsi method, correct to two decimal places.

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Background and Explanation

The Regula Falsi method (also known as the False Position method) is an iterative numerical technique for finding roots of equations. It improves upon the bisection method by using linear interpolation to estimate the root position within a bracketing interval $[a,b]$ where the function changes sign.

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Solution

First, we verify that a root exists in the interval $[2,3]$ by checking for a sign change.

Define the function: $f(x)=x^3-4x-1$

Evaluate at the endpoints: $f(2)=2^3-4(2)-1=8-8-1=-1<0$ $f(3)=3^3-4(3)-1=27-12-1=14>0$

Since $f(2)<0$ and $f(3)>0$, by the Intermediate Value Theorem, there is at least one root in $[2,3]$.

The Regula Falsi formula is: $x=\frac{af(b)-bf(a)}{f(b)-f(a)}$

First Iteration ($x_0$)

Apply the formula with $a=2$ and $b=3$:

$\begin{array}{rl}{x}_0& =\frac{2f(3)-3f(2)}{f(3)-f(2)}\\ & =\frac{2(14)-3(-1)}{14-(-1)}\\ & =\frac{28+3}{15}\\ & =\frac{31}{15}\\ & =2.0667\end{array}$

Evaluate the function at $x_0$: $\begin{array}{rl}f\left(x_0\right)& =f(2.0667)\\ & =(2.0667{)}^3-4(2.0667)-1\\ & =-0.4394<0\end{array}$

Since $f(2.0667)<0$ and $f(3)>0$, the root lies in the updated interval $[2.0667,3]$.

Second Iteration ($x_1$)

Apply the formula with $a=2.0667$ and $b=3$:

$\begin{array}{rl}{x}_1& =\frac{2.0667f(3)-3f(2.0667)}{f(3)-f(2.0667)}\\ & =\frac{2.0667(14)-3(-0.4394)}{14-(-0.4394)}\\ & =2.0951\end{array}$

Evaluate the function: $\begin{array}{rl}f\left(x_1\right)& =f(2.0951)\\ & =(2.0951{)}^3-4(2.0951)-1\\ & =-0.1841<0\end{array}$

The updated interval is $[2.0951,3]$.

Third Iteration ($x_2$)

Apply the formula with $a=2.0951$ and $b=3$:

$\begin{array}{rl}{x}_2& =\frac{2.0951f(3)-3f(2.0951)}{f(3)-f(2.0951)}\\ & =\frac{2.0951(14)-3(-0.1841)}{14-(-0.1841)}\\ & =2.1068\end{array}$

Evaluate the function: $\begin{array}{rl}f\left(x_2\right)& =f(2.1068)\\ & =(2.1068{)}^3-4(2.1068)-1\\ & =-0.0759<0\end{array}$

The updated interval is $[2.1068,3]$.

Fourth Iteration ($x_3$)

Apply the formula with $a=2.1068$ and $b=3$:

$\begin{array}{rl}{x}_3& =\frac{2.1068f(3)-3f(2.1068)}{f(3)-f(2.1068)}\\ & =\frac{2.1068(14)-3(-0.0759)}{14-(-0.0759)}\\ & =2.1116\end{array}$

Evaluate the function: $\begin{array}{rl}f\left(x_3\right)& =f(2.1116)\\ & =(2.1116{)}^3-4(2.1116)-1\\ & =-0.0311<0\end{array}$

The updated interval is $[2.1116,3]$.

Fifth Iteration ($x_4$)

Apply the formula with $a=2.1116$ and $b=3$:

$\begin{array}{rl}{x}_4& =\frac{2.1116f(3)-3f(2.1116)}{f(3)-f(2.1116)}\\ & =\frac{2.1116(14)-3(-0.0311)}{14-(-0.0311)}\\ & =2.1136\end{array}$

Convergence Check

Comparing the last two approximations:

• $x_3=2.1116\approx 2.11$ (to 2 decimal places)
• $x_4=2.1136\approx 2.11$ (to 2 decimal places)

Since both values agree when rounded to two decimal places, the iteration stops here.

Therefore, the root of the equation correct to two decimal places is $2.11$.

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Key Formulas or Methods Used

• Intermediate Value Theorem: If $f(a)\cdot f(b)<0$, then at least one root exists in $[a,b]$
• Regula Falsi Formula: $x_{\text{new}}=\frac{af(b)-bf(a)}{f(b)-f(a)}$
• Interval Update Rule:
  • If $f(x_{\text{new}})<0$, new interval is $[x_{\text{new}},b]$
  • If $f(x_{\text{new}})>0$, new interval is $[a,x_{\text{new}}]$

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Summary of Steps

1. Verify bracket: Confirm $f(2)=-1<0$ and $f(3)=14>0$, so root $\in [2,3]$
2. First iteration: Calculate $x_0=2.0667$, find $f(x_0)=-0.4394<0$, update interval to $[2.0667,3]$
3. Second iteration: Calculate $x_1=2.0951$, find $f(x_1)=-0.1841<0$, update interval to $[2.0951,3]$
4. Third iteration: Calculate $x_2=2.1068$, find $f(x_2)=-0.0759<0$, update interval to $[2.1068,3]$
5. Fourth iteration: Calculate $x_3=2.1116$, find $f(x_3)=-0.0311<0$, update interval to $[2.1116,3]$
6. Fifth iteration: Calculate $x_4=2.1136$
7. Check convergence: Both $x_3$ and $x_4$ round to $2.11$ to two decimal places, so the root is 2.11