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Example
Created on 2024-06-20Asked by Aiden Young (Solvelet student)
Use the bisection method to find an approximate root of the function f(x)=x34x9 f(x) = x^3 - 4x - 9 in the interval [2,3][2, 3].

Solution

To find an approximate root of the function f(x)=x34x9 f(x) = x^3 - 4x - 9 in the interval [2,3][2, 3] using the bisection method, follow these steps: 1. Calculate f(2) f(2) and f(3) f(3) : f(2)=23429=889=9, f(2) = 2^3 - 4 \cdot 2 - 9 = 8 - 8 - 9 = -9, f(3)=33439=27129=6. f(3) = 3^3 - 4 \cdot 3 - 9 = 27 - 12 - 9 = 6. 2. Since f(2) f(2) and f(3) f(3) have opposite signs, a root lies in the interval [2,3][2, 3]. 3. Compute the midpoint c c : c=2+32=2.5. c = \frac{2 + 3}{2} = 2.5. 4. Calculate f(2.5) f(2.5) : f(2.5)=(2.5)342.59=15.625109=3.375. f(2.5) = (2.5)^3 - 4 \cdot 2.5 - 9 = 15.625 - 10 - 9 = -3.375. 5. Since f(2.5) f(2.5) and f(3) f(3) have opposite signs, a root lies in the interval [2.5,3][2.5, 3]. 6. Repeat the process: c=2.5+32=2.75, c = \frac{2.5 + 3}{2} = 2.75, f(2.75)=(2.75)342.759=20.796875119=0.796875. f(2.75) = (2.75)^3 - 4 \cdot 2.75 - 9 = 20.796875 - 11 - 9 = 0.796875. 7. Continue the iterations until the interval is sufficiently small. Therefore, the approximate root of f(x)=x34x9 f(x) = x^3 - 4x - 9 in the interval [2,3][2, 3] is around 2.75 2.75 . Solved on Solvelet with Basic AI Model
Some of the related questions asked by Mason Walker on Solvelet
1. Use the bisection method to find a root of the equation f(x)=x32x25=0 f(x) = x^3 - 2x^2 - 5 = 0 within the interval [1,2] [1, 2] .2. Apply the secant method to solve the equation f(x)=ex2x1f(x) = e^x - 2x - 1 with initial guesses x0=0x_0 = 0 and x1=1x_1 = 1.
DefinitionNumerical methods there are algorithms used for solving mathematical methods using numerical means and iterative methods. These are necessary for numerical approximations of the solutions of the equations which are too complicated to be solved analytically.
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