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Example
Created on 2024-06-20Asked by Eleanor Flores (Solvelet student)
Use the bisection method to find the root of f(x)=x3−x−2f(x) = x^3 - x - 2 on the interval [1,2][1, 2] with tolerance ϵ=0.01\epsilon = 0.01.

Solution

To find the root of f(x)=x3−x−2f(x) = x^3 - x - 2 using the bisection method on the interval [1,2][1, 2] with tolerance ϵ=0.01\epsilon = 0.01: 1. **Check the values at the endpoints:** f(1)=13−1−2=−2, f(1) = 1^3 - 1 - 2 = -2, f(2)=23−2−2=4. f(2) = 2^3 - 2 - 2 = 4. Since f(1)⋅f(2)<0f(1) \cdot f(2) < 0, there is a root in [1,2][1, 2]. 2. **Perform the bisection method:** \begin{itemize} \item Calculate the midpoint cc of the interval [a,b][a, b]: c=a+b2. c = \frac{a + b}{2}. \item Evaluate f(c)f(c): f(c)=c3−c−2. f(c) = c^3 - c - 2. \item Determine the subinterval containing the root: \begin{itemize} \item If f(a)⋅f(c)<0f(a) \cdot f(c) < 0, the root is in [a,c][a, c]. \item If f(b)⋅f(c)<0f(b) \cdot f(c) < 0, the root is in [c,b][c, b]. \end{itemize} \item Update the interval to the new subinterval and repeat until b−a2<ϵ\frac{b - a}{2} < \epsilon. \end{itemize} 3. **Iterations:** \begin{align*} & a_0 = 1, \, b_0 = 2 \\ & c_0 = \frac{1 + 2}{2} = 1.5 \\ & f(1.5) = (1.5)^3 - 1.5 - 2 = -0.125 \\ & \text{Since } f(1.5) \cdot f(2) < 0, \, a_1 = 1.5, \, b_1 = 2 \\ & c_1 = \frac{1.5 + 2}{2} = 1.75 \\ & f(1.75) = (1.75)^3 - 1.75 - 2 = 1.609 \\ & \text{Since } f(1.5) \cdot f(1.75) < 0, \, a_2 = 1.5, \, b_2 = 1.75 \\ & c_2 = \frac{1.5 + 1.75}{2} = 1.625 \\ & f(1.625) = (1.625)^3 - 1.625 - 2 = 0.666 \\ & \text{Since } f(1.5) \cdot f(1.625) < 0, \, a_3 = 1.5, \, b_3 = 1.625 \\ & \dots \\ & \text{Continue until } \frac{b_n - a_n}{2} < 0.01. \end{align*} The approximate root is x≈1.75x \approx 1.75 (up to the specified tolerance). Solved on Solvelet with Basic AI Model
Some of the related questions asked by Jackson Gonzalez on Solvelet
1. Use the bisection method to find an approximate root of the equation f(x)=x3−5x2+4x+3 f(x) = x^3 - 5x^2 + 4x + 3 in the interval [1,3] [1, 3] .2. Apply Newton's method to find an approximate root of the equation f(x)=ex−3x2+2 f(x) = e^x - 3x^2 + 2 in the vicinity of x=1 x = 1 .,
DefinitionRoot finding methods are numerical methods used to find the roots (solutions) of equations. Some of these ways are bisections, Newton Raphson method and secant method. For example Newton-Raphson method to solve the function f(x)=0 made a single operation of the form xn+1=xn−f′(xn)​f(xn) of improving the approximation of the solution of the given root.
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