Question Statement

Find the real root of the equation $x^3-4x-9=0$ that lies in the interval $[2,3]$ using the Regula Falsi method. Continue the iterations until successive approximations agree to two decimal places.

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Background and Explanation

The Regula Falsi (False Position) method is a numerical root-finding technique that uses linear interpolation between two points $a$ and $b$ where the function changes sign. The method generates successive approximations by maintaining the bracketing property—ensuring the root remains trapped between points where $f(x)$ has opposite signs.

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Solution

Let $f(x)=x^3-4x-9$.

Verification of Root Interval

First, evaluate the function at the endpoints of $[2,3]$:

$f(2)=2^3-4(2)-9=8-8-9=-9<0$

$f(3)=3^3-4(3)-9=27-12-9=6>0$

Since $f(2)<0$ and $f(3)>0$, a root lies in the interval $[2,3]$.

First Approximation ($x_0$)

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

With $a=2$ and $b=3$:

$x_0=\frac{2f(3)-3f(2)}{f(3)-f(2)}=\frac{2(6)-3(-9)}{6-(-9)}=\frac{12+27}{15}=\frac{39}{15}=2.6$

Evaluate $f(x_0)$:

$f(2.6)=(2.6{)}^3-4(2.6)-9=17.576-10.4-9=-1.824<0$

Since $f(2.6)<0$ and $f(3)>0$, the updated interval is $[2.6,3]$.

Second Approximation ($x_1$)

Using $a=2.6$ and $b=3$:

$x_1=\frac{2.6f(3)-3f(2.6)}{f(3)-f(2.6)}=\frac{2.6(6)-3(-1.824)}{6-(-1.824)}=\frac{15.6+5.472}{7.824}=2.6933$

Evaluate $f(x_1)$:

$f(2.6933)=(2.6933{)}^3-4(2.6933)-9=-0.2364<0$

The updated interval is $[2.6933,3]$.

Third Approximation ($x_2$)

Using $a=2.6933$ and $b=3$:

$x_2=\frac{2.6933f(3)-3f(2.6933)}{f(3)-f(2.6933)}=\frac{2.6933(6)-3(-0.2364)}{6-(-0.2364)}=\frac{16.1598+0.7092}{6.2364}=2.7049$

Evaluate $f(x_2)$:

$f(2.7049)=(2.7049{)}^3-4(2.7049)-9=-0.0292<0$

The updated interval is $[2.7049,3]$.

Fourth Approximation ($x_3$)

Using $a=2.7049$ and $b=3$:

$x_3=\frac{2.7049f(3)-3f(2.7049)}{f(3)-f(2.7049)}$

$x_3=\frac{2.7049(6)-3(-0.0292)}{6-(-0.0292)}=\frac{16.2294+0.0876}{6.0292}=2.7063$

Convergence Check

Since the values of $x_2=2.7049$ and $x_3=2.7063$ agree to two decimal places (both round to $2.71$), we stop the iteration.

Therefore, correct to two decimal places, the root of the equation is $\boxed{2.71}$.

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Key Formulas or Methods Used

• Regula Falsi Formula: $x_{n+1}=\frac{af(b)-bf(a)}{f(b)-f(a)}$ where $[a,b]$ is the current bracketing interval
• Function Definition: $f(x)=x^3-4x-9$
• Interval Update Rule: If $f(x_n)<0$, the new interval becomes $[x_n,b]$; if $f(x_n)>0$, the new interval becomes $[a,x_n]$ (maintaining the sign change)

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Summary of Steps

1. Verify bracket: Check that $f(2)=-9<0$ and $f(3)=6>0$, confirming a root in $[2,3]$.
2. First iteration: Calculate $x_0=2.6$; since $f(2.6)=-1.824<0$, update interval to $[2.6,3]$.
3. Second iteration: Calculate $x_1=2.6933$; since $f(2.6933)=-0.2364<0$, update interval to $[2.6933,3]$.
4. Third iteration: Calculate $x_2=2.7049$; since $f(2.7049)=-0.0292<0$, update interval to $[2.7049,3]$.
5. Fourth iteration: Calculate $x_3=2.7063$.
6. Terminate: Since $x_2$ and $x_3$ agree to two decimal places, the root is $2.71$.