Question

Regula Falsi method is used to find a root of the equation \(f(x)=x^3-x-1\) by choosing the initial guesses \(x_0=1\) and \(x_1=2\). What would be the value of the next iteration (i.e. \(x_3\)) up to two decimal places?

• A. 1.76
• B. 1.56
• C. 1.36
• D. 1.25

Hint:

Apply the Regula Falsi (false position) formula twice: first to get x2 from x0, x1, then again from the new bracket to get x3.

Answer 1

The Correct Option is D

Step 1: Evaluate the function at the initial guesses.

\[f(x_0)=f(1)=1^3-1-1=-1, \qquad f(x_1)=f(2)=2^3-2-1=5\]

Since \(f(x_0)\) and \(f(x_1)\) have opposite signs, the root lies in \([1,2]\).

Step 2: The Regula Falsi formula for the next approximation is

\[x_2 = \frac{x_0f(x_1)-x_1f(x_0)}{f(x_1)-f(x_0)}\]

Step 3: First iteration:

\[x_2 = \frac{(1)(5)-(2)(-1)}{5-(-1)} = \frac{5+2}{6} = \frac{7}{6} \approx 1.1667\]

Evaluate \(f(x_2)\):

\[f(1.1667) = (1.1667)^3-1.1667-1 \approx 1.5880-1.1667-1 = -0.5787\]

Step 4: Since \(f(x_2)\) has the same sign as \(f(x_0)\) (both negative), the new bracket is \([x_2,x_1]=[1.1667,2]\), and \(x_1=2\) is retained for the second iteration.

Step 5: Second iteration, using \(a=x_2=1.1667\), \(f(a)=-0.5787\), \(b=x_1=2\), \(f(b)=5\):

\[x_3 = \frac{af(b)-bf(a)}{f(b)-f(a)} = \frac{(1.1667)(5)-(2)(-0.5787)}{5-(-0.5787)}\]

Step 6: Simplify the numerator and denominator:

\[x_3 = \frac{5.8333+1.1574}{5.5787} = \frac{6.9907}{5.5787} \approx 1.2531\]

Step 7: Rounding to two decimal places:

\[\boxed{x_3 \approx 1.25}\]